10. (20 points) Given that the probability of forming an obtuse triangle by randomly selecting three vertices from a regular $n$-sided polygon is $\frac{93}{125}$, find all possible values of the positive integer $n$
10. (1) When $n$ is even, let $n=2 k$, and denote the regular $2 k$-gon as $A_{-k+1} A_{-k+2} \cdots A_{k}$.

Without loss of generality, choose vertex $A_{0}$. If we also choose its diametrically opposite point $A_{k}$, then the resulting triangles are all right triangles. In fact, when $A_{0}$ is fixed, choosing any two other points to form an obtuse triangle with $A_{0}$ is equivalent to the difference in the indices of the chosen points being no more than $k-1$. If the indices of these two points have the same sign, the condition is always satisfied, giving a total of $2 \mathrm{C}_{k-1}^{2}$ ways. If we choose point $A_{j}(1 \leqslant j \leqslant k-1)$, then there are exactly $k-1-j$ vertices with negative indices that form an obtuse triangle with $A_{0}$ and $A_{j}$, which gives
$$
\sum_{j=1}^{k-1}(k-1-j)=\sum_{i=0}^{k-2} i=\mathrm{C}_{k-1}^{2}
$$

ways.

Thus, we get a total of $3 \mathrm{C}_{k-1}^{2}$ obtuse triangles. There are $\mathrm{C}_{2 k-1}^{2}$ ways to form a triangle with $A_{0}$, so the probability that the resulting triangle is an obtuse triangle is
$$
\frac{3 \mathrm{C}_{k-1}^{2}}{\mathrm{C}_{2 k-1}^{2}}=\frac{\frac{3(k-1)(k-2)}{2}}{\frac{(2 k-1)(2 k-2)}{2}}=\frac{3(k-2)}{2(2 k-1)}=\frac{93}{125} .
$$

Solving this, we get $k=188, n=376$.
(2) When $n$ is odd, let $n=2 k+1$, and denote the regular $2 k+1$-gon as $A_{-k} A_{-k+1} \cdots A_{k}$.
The number of ways to choose vertices to form a triangle with $A_{0}$ is $\mathrm{C}_{2 k}^{2}$. Similarly, the number of obtuse triangles formed by $m$ $(m=0,1,2)$ vertices with positive indices is $\mathrm{C}_{k}^{2}$. Therefore, the probability that the resulting triangle is an obtuse triangle is
$$
\frac{3 \mathrm{C}_{k}^{2}}{\mathrm{C}_{2 k}^{2}}=\frac{\frac{3 k(k-1)}{2}}{\frac{2 k(2 k-1)}{2}}=\frac{3(k-1)}{2(2 k-1)}=\frac{93}{125} .
$$

Solving this, we get $k=63, n=127$.
In summary, the possible values of the positive integer $n$ are 376 and 127.