Let $n$ be a positive integer. We are given a regular $4n$-gon in the plane. We divide its vertices in $2n$ pairs and connect the two vertices in each pair by a line segment. What is the maximal possible amount of distinct intersections of the segments?
1. **Understanding the Problem:**
   We are given a regular $4n$-gon and need to find the maximum number of distinct intersections of line segments formed by pairing its vertices into $2n$ pairs.

2. **Initial Setup:**
   Let the vertices of the $4n$-gon be labeled as $V_1, V_2, \ldots, V_{4n}$ in clockwise order. We will pair these vertices and connect each pair with a line segment (diagonal).

3. **Counting Intersections:**
   Each pair of diagonals can intersect at most once. We need to count the number of such intersections. 

4. **Choosing Pairs:**
   Consider the diagonals formed by pairing vertices such that each pair $(V_i, V_{i+2n})$ for $i = 1, 2, \ldots, 2n$. This ensures that each diagonal spans half the polygon, maximizing the potential for intersections.

5. **Calculating Intersections:**
   To find the number of intersections, we need to count how many ways we can choose 2 pairs of diagonals that intersect. Each pair of diagonals $(d_i, d_j)$ intersects if and only if they cross each other inside the polygon.

6. **Using Combinatorics:**
   The number of ways to choose 2 pairs from $2n$ pairs is given by the binomial coefficient:
   \[
   \binom{2n}{2} = \frac{2n(2n-1)}{2} = n(2n-1)
   \]

7. **Adjusting for Overcounting:**
   We need to ensure that no three diagonals intersect at a single point. In a regular polygon, no three diagonals can intersect at a single point unless they all pass through the center, which is not the case here.

8. **Final Calculation:**
   The maximum number of distinct intersections is given by:
   \[
   n(2n-1)
   \]

The final answer is $\boxed{2n^2 - n}$.