Suppose there are $n$ points on the plane, no three of which are collinear. Draw $n-1$ non-intersecting segments (except possibly at endpoints)  between pairs of points, such that it is possible to travel between any two points by travelling along the segments. Such a configuration of points and segments is called a [i]network[/i]. Given a network, we may assign labels from $1$ to $n-1$ to each segment such that each segment gets a different label. Define a [i]spin[/i] as the following operation:

    $\bullet$ Choose a point $v$ and rotate the labels of its adjacent segments clockwise. Formally, let $e_1,e_2,\cdots,e_k$ be the segments which contain $v$ as an endpoint, sorted in clockwise order (it does not matter which segment we choose as $e_1$). Then, the label of $e_{i+1}$ is replaced with the label of $e_{i}$ simultaneously for all $1 \le i \le k$. (where $e_{k+1}=e_{1}$)

A network is [i]nontrivial[/i] if there exists at least $2$ points with at least $2$ adjacent segments each. A network is [i]versatile[/i] if any labeling of its segments can be obtained from any initial labeling using a finite amount of spins. Find all integers $n \ge 5$ such that any nontrivial network with $n$ points is versatile.

[i]Proposed by Yeoh Zi Song[/i]
1. **Interpret the Network as a Tree**: We start by interpreting the network as a tree and root it at an arbitrary vertex. For a permutation \( p_1, p_2, \cdots, p_n \) of \(\{1, 2, \cdots, n\}\), we say that it is even if it can be transformed into the identity permutation with an even number of transpositions, and odd otherwise. This is well-defined and can be proven by considering the parity of the number of cycles in the graph \( i \rightarrow p_i \).

2. **Define Even and Odd Labeling**: Fix any target labeling of edges in the tree. We say that our current labeling is even if we can transform it into the target labeling by an even number of label swaps (where we are allowed to swap *any* two labels). Call the labeling odd otherwise.

3. **Key Claim**: Assume that our network is nontrivial. The key claim is that the current labeling cannot reach the target labeling using a finite number of spins if and only if:
   - All vertices have odd degree.
   - The current labeling of edges is odd.

4. **Implication of the Claim**: This claim implies the result because if a nontrivial network is not versatile, then it must have an even number of vertices (or else the sum of degrees is odd). For any even \( n \geq 6 \), we can construct a nonversatile nontrivial network by taking a tree where the root has \( n-3 \) children, and one of its children has 2 children. It is impossible to swap exactly one pair of labels in the network.

5. **Proof of the Claim**:
   - **Condition Satisfaction**: Firstly, we prove that if the conditions are satisfied, then the target labeling can never be reached. Since every vertex has odd degree, every spin is equivalent to some even number of transpositions. Thus, the parity of the network is always preserved after a spin. This implies that we can never reach the target labeling using spins, since we begin with an odd labeling.
   - **Induction on \( n \)**: We will use induction on \( n \), with the base case \( n=4 \) being trivial because the only valid nontrivial network is isomorphic to a path, and we can swap any two adjacent labels in a path (which allows us to reach any permutation of labels).

6. **Inductive Step**:
   - **Step 1**: Make \( S_i \) contain only edge labels that belong to \( S_i \) in the target labeling.
     - Let \( m \) be the number of edge labels that are currently in the wrong subtree. We will show that if \( m > 0 \), then there exists a sequence of spins that decreases \( m \).
     - Choose an edge \( e \in S_i \) such that its current label \( l_e \) belongs in another subtree \( S_j \) in the target labeling. Choose any label \( l_f \) that is currently on some edge \( f \in S_j \) but does not belong in \( S_j \) in the target labeling. We will move \( l_e \) to \( S_j \) and \( l_f \) to \( S_i \) while ensuring that every other label stays in its subtree.
     - **Lemma 1**: Let \( T \) be a tree rooted at vertex \( r \). Let \( v_1, v_2, \cdots, v_k \) be the children of \( r \). Suppose not all vertices \( v_i \) have degree 1. Then, it is possible to swap the labels on \( (r, v_i) \) and \( (r, v_j) \) for any \( i < j \) while ensuring that every other label stays in its own subtree.
       - **Proof**: Without loss of generality, assume that \( v_1 \) has degree \( \geq 2 \). The following sequence of operations suffice:
         - Spin \( r \) clockwise \( i-1 \) times.
         - Spin \( v_1 \) clockwise once.
         - Spin \( r \) clockwise \( j-i \) times.
         - Spin \( v_1 \) counterclockwise once.
         - Spin \( r \) counterclockwise \( j-i \) times.
         - Spin \( v_1 \) clockwise once.
         - Spin \( r \) counterclockwise \( i-1 \) times.
       - To finish this step, we just have to move \( l_e \) to the parent edge of \( c_i \) by repeatedly spinning the label up the tree, and similarly move \( l_f \) to the parent edge of \( c_j \). Then, apply Lemma 1 to swap the labels of the parent edges of \( c_i \) and \( c_j \). This shows that we can decrease \( m \) and eventually move each label in the correct subtree, as desired.

   - **Step 2**: Make all subtrees \( S_i \) have a labeling of even parity.
     - **Lemma 2**: Let \( T \) be a tree rooted at vertex \( r \). Let \( v_1, v_2, \cdots, v_k \) be the children of \( r \) and the corresponding edge labels be \( l_1, l_2, \cdots, l_k \). Suppose \( v_1 \) has degree at least 2 and its leftmost child is \( w_1 \) with edge label \( w \). Then, it is possible to swap the edge labels \( l_1 \) and \( w \) while cyclically shifting \( l_2, \cdots, l_k \) to the left while leaving other labels intact.
       - **Proof**: The following sequence of operations suffice:
         - Spin \( v_1 \) clockwise once.
         - Spin \( r \) clockwise once.
         - Spin \( v_1 \) counterclockwise once.
         - Spin \( r \) counterclockwise once.
         - Spin \( v_1 \) clockwise once.
         - Spin \( r \) clockwise once.
         - Spin \( v_1 \) counterclockwise once.
     - **Corollary**: If the root \( r \) has even degree (say \( k \)), then we can always make the parity of the labeling of each subtree \( S_i \) even.
       - **Proof**: Without loss of generality, suppose \( i=1 \). If \( S_1 \) has \( \leq 1 \) edges, its labeling is clearly even. Now, suppose the labeling in \( S_1 \) is currently odd. Apply Lemma 2 \( k-1 \) times. Since \( k-1 \) is odd, the edge labels \( l_1 \) and \( w \) in the lemma are swapped. However, all the other labels are left intact (because cyclically shifting \( k-1 \) times changes nothing). This swaps a pair of labels in \( S_1 \) which makes its new labeling even without affecting any other labels.
     - Hence, we may assume \( r \) has odd degree. Let \( C_1, C_2, \cdots, C_m \) be the subtrees among \( S_i \) that currently have an odd labeling. If \( m \) is odd, then at least one non-root vertex of the tree has even degree (or else this implies every vertex has odd degree and the labeling of the entire tree is odd). Performing a spin on that vertex will make \( m \) even. Hence, we assume that \( m \) is even.
     - To finish this step, we claim that we can make the labelings of \( C_1 \) and \( C_2 \) even without affecting labels outside these subtrees. Clearly, \( C_1, C_2 \) has \( \geq 2 \) edges (or else it already has even labeling). Without loss of generality, assume that \( C_1 = S_1 \) and \( C_2 = S_i \) for some \( i > 1 \). Let us apply the following series of operations:
       - Apply Lemma 2.
       - Spin \( r \) counterclockwise once.
       - Apply Lemma 2, with \( S_i \) now treated as the leftmost subtree (i.e., we swap the edge labels of two of \( v_i \)'s adjacent edges).
     - Notice that after these operations, the parity of \( C_1 \) and \( C_2 \) are flipped. Additionally, the labels of \( (r, c_2), (r, c_3), \cdots, (r, c_{i-1}), (r, c_{i+1}), \cdots, (r, c_{k}) \) are cyclically shifted. However, since \( k-2 \) is odd, we can repeat this operation \( k-2 \) times and ensure that the parity of \( C_1 \) and \( C_2 \) is even while keeping all labels in the correct subtrees. This completes the proof of this step.

   - **Step 3**: If \( S_i \) is a *trivial* network, ensure all its labels are correct.
     - Note that \( S_i \) has an even labeling after Step 2. If \( S_i \) has \( \leq 1 \) edges, we are done, so assume \( S_i \) has \( \geq 2 \) edges. We may assume without loss of generality \( i=1 \). Let \( v \) be the child of \( r \) in \( S_1 \). Let the labels on the edges adjacent to \( v \) be \( e_1, e_2, \cdots, e_m \) (in clockwise order). The key observation is that we can swap \( e_i \) and \( e_{i+1} \) by applying Lemma 2 (after spinning \( v \) to make \( e_i \) the parent edge), at the cost of cyclically shifting the labels on the other \( k-1 \) children of \( r \) once.
     - Firstly, use this operation to swap all the labels \( e_i \) into their correct positions. Suppose we used this operation \( t_1 \) times. Note that \( t_1 \) is even, since \( S_i \) has an even labeling. Now, we apply Lemma 2 \( 2t_2 \) times where \( t_2 \) is some positive integer satisfying \( k-1 \mid t_1 + 2t_2 \). Observe that \( t_2 \) is guaranteed to exist because \( t_1 \) is even. After these operations, all the edge labels around \( r \) are in the correct subtrees, and the subtree \( S_1 \) is solved completely. This completes the last step.

7. **Conclusion**: After performing all three steps, we can now safely apply the induction hypothesis on all the nontrivial \( S_i \)'s and obtain the target labeling, as desired. This proves the inductive step and completes the proof. 

\(\blacksquare\)

The final answer is all odd \( \boxed{ n } \).