Problem 1639

### Part 1: Maximum Value of $I$

1. Understanding the Problem:
- We have $n$ distinct points on a circumference.
- We connect each point to the $m$-th point counterclockwise.
- We continue this process until no new segment is constructed.
- We need to find the maximum number of intersections $I$ inside the circle.

2. Counting Intersections:
- Each segment can intersect with other segments.
- For each of the $n$ diagonals, there are $2(m-1)$ potential intersection points with other diagonals.
- Each intersection point is counted twice (once for each segment involved in the intersection).

3. Maximum Intersections:
- The maximum number of intersections $I$ can be given by:
$$I \leq (m-1)n$$
- This is because each of the $n$ diagonals can intersect with $2(m-1)$ other diagonals.

### Part 2: Proving $I \geq n$

1. Injection Argument:
- Consider each diagonal $A_iA_{i+m}$.
- For each diagonal, consider the point of intersection between it and $A_{i+m-1}A_{i+2m-1}$.
- This gives an injection from the set of $n$ diagonals to the set of intersection points $I$.

2. Conclusion:
- Since there is an injection from the set of $n$ diagonals to $I$, we have:
$$I \geq n$$

### Special Case: $m = 3$ and $n$ is an Even Number

1. Construction:
- Start with a regular $\frac{n}{2}$-gon $X_0, X_2, \ldots, X_{n-2}$.
- Take $X_1, X_3, \ldots, X_{n-1}$ to be the midpoints of $X_0X_2, X_2X_4, \ldots, X_{n-2}X_0$ respectively.
- Define $A_k = X_kX_{k+2} \cap X_{k+3}X_{k+5}$ for all $k \in \{0, 1, \ldots, n-1\}$ (indices modulo $n$).

2. Verification:
- This construction ensures that each diagonal intersects exactly once with another diagonal.
- Therefore, the number of intersections $I$ is exactly $n$.

The final answer is $\boxed{ I = (m-1)n }$ and $I \geq n$ with $I = n$ for $m = 3$ and $n$ even.