Let $ n\geq 2$ be a positive integer. Find the [u]maximum[/u] number of segments with lenghts greater than $ 1,$ determined by $ n$ points which lie on a closed disc with radius $ 1.$
1. **Projection of Points onto the Circle:**
   - Without loss of generality (WLOG), we can assume that the \( n \) points are on the boundary of the circle. This is because projecting all points onto the boundary does not change the lengths of the segments.

2. **Condition for Segment Length Greater than 1:**
   - A segment between two points on the circle has a length greater than 1 if and only if the central angle between the two points is greater than \( \pi/3 \). This is because the chord length \( l \) of a circle with radius \( r \) and central angle \( \theta \) is given by:
     \[
     l = 2r \sin\left(\frac{\theta}{2}\right)
     \]
     For \( r = 1 \) and \( l > 1 \):
     \[
     2 \sin\left(\frac{\theta}{2}\right) > 1 \implies \sin\left(\frac{\theta}{2}\right) > \frac{1}{2} \implies \frac{\theta}{2} > \frac{\pi}{6} \implies \theta > \frac{\pi}{3}
     \]

3. **Graph Representation:**
   - Consider a graph \( G \) where each vertex represents a point on the circle, and an edge exists between two vertices if the segment between the corresponding points is greater than 1. This means the edge exists if the central angle between the points is greater than \( \pi/3 \).

4. **Application of Turán's Theorem:**
   - We need to find the maximum number of edges in this graph \( G \) such that no subset of 6 vertices forms a complete subgraph \( K_6 \). This is because if 6 points form a \( K_6 \), then at least one of the segments will have a central angle less than or equal to \( \pi/3 \), contradicting our condition.
   - Turán's theorem states that the maximum number of edges in a graph with \( n \) vertices that does not contain \( K_{r+1} \) as a subgraph is given by:
     \[
     \left(1 - \frac{1}{r}\right) \frac{n^2}{2}
     \]
     For \( r = 5 \) (since we are avoiding \( K_6 \)):
     \[
     \left(1 - \frac{1}{5}\right) \frac{n^2}{2} = \frac{4}{5} \cdot \frac{n^2}{2} = \frac{2n^2}{5}
     \]

5. **Conclusion:**
   - Therefore, the maximum number of segments with lengths greater than 1, determined by \( n \) points on a closed disc with radius 1, is:
     \[
     \boxed{\frac{2n^2}{5}}
     \]