Let $ n$ be an integer greater than $ 3.$ Points $ V_{1},V_{2},...,V_{n},$ with no three collinear, lie on a plane. Some of the segments $ V_{i}V_{j},$ with $ 1 \le i < j \le n,$ are constructed. Points  $ V_{i}$ and $ V_{j}$ are [i]neighbors[/i] if  $ V_{i}V_{j}$ is constructed. Initially, chess pieces $ C_{1},C_{2},...,C_{n}$ are placed at points $ V_{1},V_{2},...,V_{n}$ (not necessarily in that order) with exactly one piece at each point. In a move, one can choose some of the $ n$ chess pieces, and simultaneously relocate each of the chosen piece from its current position to one of its neighboring positions such that after the move, exactly one chess piece is at each point and no two chess pieces have exchanged their positions. A set of constructed segments is called [i]harmonic[/i] if for any initial positions of the chess pieces, each chess piece $ C_{i}(1 \le i \le n)$ is at the point $ V_{i}$ after a finite number of moves. Determine the minimum number of segments in a harmonic set.
1. **Understanding the Problem:**
   We need to determine the minimum number of segments in a harmonic set such that for any initial positions of the chess pieces, each chess piece \( C_i \) (for \( 1 \le i \le n \)) is at the point \( V_i \) after a finite number of moves. 

2. **Graph Representation:**
   Represent the points \( V_1, V_2, \ldots, V_n \) as vertices of a graph, and the segments \( V_iV_j \) as edges. The problem then translates to finding the minimum number of edges such that the graph is harmonic.

3. **Cycle and Permutations:**
   If we have \( n \) segments, we can form a single cycle. However, a single cycle can only generate \( n \) different permutations, but we need to be able to reach all \( n! \) permutations.

4. **Adding an Extra Edge:**
   Suppose we have \( n+1 \) segments. Construct the segments so that \( V_i \) is connected to \( V_{i+1} \) (including \( V_n \) to \( V_1 \)), forming an \( n \)-cycle. Additionally, connect \( V_1 \) to \( V_4 \), forming a 4-cycle within the \( n \)-cycle.

5. **Permutations Using Cycles:**
   - By using the \( n \)-cycle, we can shift all pieces cyclically.
   - By using the 4-cycle, we can permute any 4 consecutive pieces.
   - For example, if we have 5 adjacent pieces \( a, b, c, d, e \), we can perform the following permutations:
     \[
     (a, b, c, d, e) \rightarrow (c, d, a, b, e) \rightarrow (c, a, b, e, d) \rightarrow (a, b, e, c, d)
     \]
     This shows that we can generate any 3-cycle within the 5 pieces.

6. **Achieving the Desired Permutation:**
   - Using the 3-cycle and 4-cycle, we can move \( C_1 \) to \( V_1 \), then \( C_2 \) to \( V_2 \), and so on, without disturbing the already placed pieces.
   - For the last few pieces, if we end up with \( C_{n-3}, C_{n-2}, C_n, C_{n-1} \) in \( V_{n-3}, V_{n-2}, V_{n-1}, V_n \) respectively, we can use the 4-cycle to rearrange them as needed.

7. **Final Adjustment:**
   - Use the 4-cycle to get \( C_{n-1}, C_{n-3}, C_{n-2}, C_n \).
   - Then use the 3-cycle to get \( C_{n-3}, C_{n-2}, C_{n-1}, C_n \).

Thus, the minimum number of segments in a harmonic set is \( n+1 \).

The final answer is \(\boxed{n+1}\).