Solution: If $n$ is odd, then $m = n - 3$; if $n$ is even, then $m = n - 2$.

1. First we prove the case where $n$ is odd.

a. Claim. No two diagonals within a regular $n$-gon are perpendicular.
Suppose diagonals $\overline{AB}$ and $\overline{CD}$ are perpendicular. Draw the perpendicular bisector of $AB$, which meets the regular $n$-gon at vertex $E$ (note that since $n$ is odd, $E$ exists). Let $E'$ be the point symmetric to $E$ on the regular $n$-gon:
- Note that since $n$ is odd, $E'$ cannot be a vertex of the regular $n$-gon.
- However, by symmetry, it is easy to see that $\overline{EC} = \overline{E'D}$. This means $E'$ must be a vertex of the regular $n$-gon.
Contradiction!

b. By the above Claim and the problem's assumption, these $m$ diagonals must not intersect inside the polygon; in other words, they must all emanate from the same endpoint, and there are $n-3$ diagonals emanating from the same endpoint, which proves the claim.

(a) We first prove a stronger result by induction: for all $n$, if we can select $m$ diagonals satisfying the problem's condition from a cyclic $n$-gon, then $m \le 2$. This is obvious when $n = 3$. Now, suppose $m \le n-2$ holds for $n = 1, 2, \dots, N-1$; then when $n = N$, we discuss two cases:

Case 1. Among these $m$ diagonals, there is one that does not intersect any other diagonal inside the polygon. This diagonal divides the regular $n$-gon into an $n_1$-gon and an $n_2$-gon ($n_1 + n_2 = n + 2$). Without loss of generality, suppose the $n_1$-gon lies entirely on one side of some diameter of the circumscribed circle $\Gamma$ of the regular $n$-gon; then it is easy to see that no two diagonals within this $n_1$-gon can be perpendicular, so within the $n_1$-gon we can only choose non-intersecting diagonals. As in the discussion of 1.(a) above, the maximum possible number of these is $n_1 - 3$.
On the other hand, by the induction hypothesis, at most $m(n_2) \le n_2 - 2$ diagonals can be chosen within the $n_2$-gon, so in total at most $(n_1 - 3) + (n_2 - 2) + 1 = n - 2$ diagonals can be chosen.

Case 2. Every one of these $m$ diagonals is perpendicular to at least one other. Let $d_1$ and $d_2$ be a pair of perpendicular diagonals; then they divide the circle into four regions:


- If there is another diagonal $d'$ that is not perpendicular to either $d_1$ or $d_2$, then its two endpoints must both belong to the same region (otherwise it would intersect $d_1$ or $d_2$).
- If we take two other intersecting diagonals $d'_1$ and $d'_2$, neither of which is perpendicular to $d_1$ or $d_2$, then their endpoints must both lie in the same region. However, it is easy to see that if the endpoints of two diagonals both lie in the same region, they cannot be perpendicular (as shown in the figure), a contradiction.

- Moreover, based on our assumption that every one of these $m$ diagonals is perpendicular to at least one other, these $m$ diagonals can only be parallel to $d_1$ or $d_2$.
Now, let $d_1^*$ be the longest diagonal among these $m$ diagonals that is parallel to $d_1$; it is easy to see that none of the other $m-1$ diagonals can share an endpoint with $d_1^*$. Similarly, let $d_2^*$ be the longest diagonal among these $m$ diagonals that is parallel to $d_2$; it is easy to see that none of the other $m-1$ diagonals can share an endpoint with $d_2^*$. Excluding the four endpoints of $d_1^*$ and $d_2^*$, each of the remaining $n-4$ vertices can be the endpoint of at most two diagonals (since these diagonals must all be parallel to $d_1$ or $d_2$), so at most
$$\frac{1}{2}(2(n-4)) + 2 = n - 2$$
diagonals can be chosen. This proves the claim.

(b) Finally, we construct a selection achieving $m = n - 2$: select two points $A$ and $A'$ of the regular $n$-gon. Choose:
- $\overline{AA'}$;
- Let $B$ and $B'$ be the two vertices to the left and right of $A'$. Choose $\overline{AB}$, $\overline{AB'}$ and $\overline{BB'}$, where $\overline{BB'}$ is perpendicular to $\overline{AA'}$.
- Let $C$ and $C'$ be the two vertices to the left and right of $A$. For all remaining vertices
$$D \notin \{A, A', B, B', C, C'\},$$
choose $\overline{AD}$.
In total there are $1+3+(n-2) = n-2$ diagonals.