[b] Problem 6.[/b] Let $m\geq 5$ and $n$ are given natural numbers, and $M$ is regular $2n+1$-gon. Find the number of the convex $m$-gons with vertices among the vertices of $M$, who have at least one acute angle.

[i]Alexandar Ivanov[/i]
To solve this problem, we need to count the number of convex \( m \)-gons with vertices among the vertices of a regular \( (2n+1) \)-gon \( M \) that have at least one acute angle. We will follow the steps outlined in the initial solution and fill in the missing details.

1. **Understanding the Problem:**
   - We have a regular \( (2n+1) \)-gon \( M \).
   - We need to count the number of convex \( m \)-gons with vertices among the vertices of \( M \) that have at least one acute angle.
   - An acute angle in a regular polygon occurs when the gap between the vertices adjacent to the acute angle is greater than \( n \).

2. **Counting Acute Angles:**
   - Fix a location for one vertex \( e \).
   - At most one of the \( n \) vertices preceding \( e \) can be used to form an acute angle.
   - This leads to a total of \( (2n+1) \left( \binom{n}{m-1} + n \binom{n}{m-2} \right) \) acute angles.

3. **Adjusting for Overcounting:**
   - Some polygons may have more than one acute angle.
   - A polygon can never have two nonadjacent acute angles because the two jumps of length \( n+1 \) would overlap.
   - We need to subtract the number of polygons with two adjacent acute angles.

4. **Counting Polygons with Two Adjacent Acute Angles:**
   - Assume the first angle starts at vertex 1.
   - We need to count the number of ways to place the remaining \( m-1 \) vertices such that:
     1. Vertex #3 is placed in position \( n+2 \) or later.
     2. If vertex #2 is placed in position \( x \), vertex #4 is placed in position \( x+n+1 \) or later.

5. **Calculating the Overcounting:**
   - Let \( x \) be the location of vertex 2 (between 2 and \( n \)).
   - Let \( y \) be the location of vertex 4 (between \( x+n+1 \) and \( 2n+1 \)).
   - We have \( y-x-1 \) choices for the placement of vertex 3.
   - We have \( \binom{2n+1-y}{m-4} \) choices for the placement of the remaining vertices.
   - The total overcounting is:
     \[
     (2n+1) \sum_{x=2}^{n} \sum_{y=x+n+1}^{2n+1} (y-x-1) \binom{2n+1-y}{m-4}
     \]

6. **Evaluating the Sum:**
   - The sum can be evaluated by breaking it down into simpler parts and using combinatorial identities.
   - However, the exact evaluation of this sum is complex and may require advanced techniques or software tools.

The final answer is \( \boxed{ (2n+1) \left( \binom{n}{m-1} + n \binom{n}{m-2} \right) - (2n+1) \sum_{x=2}^{n} \sum_{y=x+n+1}^{2n+1} (y-x-1) \binom{2n+1-y}{m-4} } \).