Let integers $n \geqslant m \geqslant 5, A_{1} A_{2} \cdots A_{2 n+1}$ be a regular $(2 n+1)$-gon, $P=\left\{A_{1}, A_{2}, \cdots, A_{2 n+1}\right\}$. Find the number of convex $m$-gons with vertices in $P$ and at least one acute interior angle.
First, we prove a lemma.

Lemma: A convex $m$-gon with vertices in $P$ has at most two acute angles, and if there are two acute angles, they must be adjacent.

Proof of the lemma: Let this convex $m$-gon be $P_{1} P_{2} \cdots P_{m}$. We only consider the case where there is at least one acute angle. Assume $\angle P_{m} P_{1} P_{2} < \frac{\pi}{2} \quad (3 \leqslant j \leqslant m-1)$,

then
$$
\angle P_{j-1} P_{j} P_{j+1} > \frac{\pi}{2} \quad (3 \leqslant j \leqslant m-1).
$$

And $\angle P_{1} P_{2} P_{3} + \angle P_{m-1} P_{m} P_{1} > \pi$, so at most one of them is an acute angle. This proves the lemma.
Returning to the original problem.
By the lemma, if a convex $m$-gon has exactly two acute angles, then the corresponding vertices are adjacent.
Assume the perimeter of the regular $(2n+1)$-gon $A_{1} A_{2} \cdots A_{2n+1}$ is $2n+1$. In the clockwise direction, the first acute angle vertex has $2n+1$ choices. Once the first acute angle vertex is chosen, starting from this point, set the arc lengths between adjacent vertices of the convex $m$-gon as $x_{1}, x_{2}, \cdots, x_{m}$, then
$$
x_{1} + x_{2} + \cdots + x_{m} = 2n+1,
$$

and $x_{1} + x_{m} \geqslant n+1, x_{m-1} + x_{m} \leqslant n$.
Let $x_{m} = n - k (1 \leqslant k \leqslant n-1)$, then the number of such convex $m$-gons is equal to the number of positive integer solutions of the indeterminate equation
$$
x_{1} + x_{2} + \cdots + x_{m-1} = n + k + 1
$$

satisfying the conditions $x_{m-1} \leqslant k, x_{1} \geqslant k+1$. Let $y_{1} = x_{1} - k, y_{i} = x_{i} (i=2,3, \cdots, m-1)$, then the number of such convex $m$-gons is equal to the number of positive integer solutions of the indeterminate equation
$$
y_{1} + y_{2} + \cdots + y_{m-1} = n + 1
$$

satisfying the condition $y_{m-1} \leqslant k$. By the exclusion method, the number of such positive integer solutions is $\mathrm{C}_{n}^{m-2} - \mathrm{C}_{n-k}^{m-2}$. Therefore, the number of convex $m$-gons with the chosen vertex as the first acute angle vertex is
$$
\sum_{k=1}^{n-1} \left( \mathrm{C}_{n}^{m-2} - \mathrm{C}_{n-k}^{m-2} \right) = (n-1) \mathrm{C}_{n}^{m-2} - \mathrm{C}_{n}^{m-1} = n \mathrm{C}_{n}^{m-2} - \mathrm{C}_{n+1}^{m-1}.
$$

Thus, the number of such convex $m$-gons is $(2n+1) \left( n \mathrm{C}_{n}^{m-2} - \mathrm{C}_{n+1}^{m-1} \right)$.