Given are $n^2$ points in the plane, such that no three of them are collinear, where $n \geq 4$ is the positive integer of the form $3k+1$. What is the minimal number of connecting segments among the points, such that for each $n$-plet of points we can find four points, which are all connected to each other?

[i]Proposed by Alexander Ivanov and Emil Kolev[/i]
1. **Understanding the Problem:**
   We are given \( n^2 \) points in the plane such that no three of them are collinear, where \( n \geq 4 \) is a positive integer of the form \( 3k + 1 \). We need to find the minimal number of connecting segments among the points such that for each \( n \)-plet of points, we can find four points that are all connected to each other.

2. **Graph Theory Approach:**
   We can model this problem using graph theory. Each point corresponds to a vertex, and each connecting segment corresponds to an edge. We need to ensure that for any subset of \( n \) vertices, there exists a \( K_4 \) (a complete subgraph on 4 vertices).

3. **Turán's Graph \( T(n,3) \):**
   Consider the Turán graph \( T(n,3) \), which is the complete tripartite graph \( K_{k,k,k+1} \) where \( n = 3k + 1 \). This graph has:
   \[
   t(n,3) = 3k^2 + 2k \text{ edges}
   \]

4. **Complete Graph \( K_n \):**
   The complete graph \( K_n \) on \( 3k + 1 \) vertices has:
   \[
   k_n = \frac{(3k+1)(3k)}{2} = \frac{9k^2 + 3k}{2} \text{ edges}
   \]

5. **Combining Graphs:**
   Consider the graph \( G = T(n,3) \cup K_n \). This graph has:
   \[
   m = t(n,3) + n^2 + k_n \text{ edges}
   \]

6. **Property Verification:**
   The graph \( G \) does not have the required property. However, any graph on \( n^2 \) vertices with more than \( m \) edges will have the property. This is because any subgraph of order \( n \) will have more than \( t(n,3) \) edges, thus containing a \( K_4 \).

7. **Conclusion:**
   The minimal number of connecting segments (edges) required is \( m + 1 \).

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