The maximum number of acute-angled triangles is $\frac{n(n-1)(n+1)}{24}$ if $n$ is odd and is $\frac{n(n-2)(n+2)}{24}$ if $n$ is even.

We first count the number $N_1$ of non-acute triangles with $A_1$ as one vertex such that $\angle A_1 < 90^\circ$. Let $\ell$ be the diameter of the circle which passes through $A_1$. Then for points $B, C$ on the circle, $\triangle A_1BC$ is not acute with $\angle BA_1C < 90^\circ$ if and only if $B$ and $C$ lie on the same side of $\ell$ (possibly on $\ell$). Suppose there are $a$ points on one side of $\ell$ and $b$ points on the other side of $\ell$ other than $A_1$. Note that the two sides share at most one common point which is the other intersection of $\ell$ and the circle. Therefore, we must have $a+b=n-1$ or $a+b=n$. We find that
$$N_1 = \binom{a}{2} + \binom{b}{2}.$$
We have to minimize $N_1$. Note that $b \ge n-1-a$ so that $\binom{b}{2} \ge \binom{n-1-a}{2}$.
WLOG assume $a \ge n-1-a$. If $n$ is odd, then $(a, n-1-a) > \left(\frac{n-1}{2}, \frac{n-1}{2}\right)$.
Applying the majorization inequality, we obtain
$$N_1 \ge \binom{a}{2} + \binom{n-1-a}{2} \ge \left(\frac{n-1}{2}\right) + \left(\frac{n-1}{2}\right) = \frac{(n-1)(n-3)}{4}.$$
The same bound holds if we replace $A_1$ by any other vertex $A_j$. We sum up all the lower bounds. Note that each non-acute triangle is counted twice in

this sum (one corresponding to each acute angle). Therefore, the number of non-acute triangles is at least
$$n \cdot \frac{(n-1)(n-3)}{4} \cdot \frac{1}{2} = \frac{n(n-1)(n-3)}{8}.$$
Thus, the number of acute triangles is at most
$$\binom{n}{3} - \frac{n(n-1)(n-3)}{8} = \frac{n(n-1)(n+1)}{24}.$$
This can be attained when $A_1A_2\cdots A_n$ is a regular polygon (since in that case all inequalities used become equalities).

Similarly, if $n$ is even, then $(a, n-1-a) > \left(\frac{n}{2}, \frac{n-2}{2}\right)$. Applying the majorization inequality, we obtain
$$N_1 \ge \binom{a}{2} + \binom{n-1-a}{2} \ge \binom{\frac{n}{2}}{2} + \binom{\frac{n-2}{2}}{2} = \frac{(n-2)^2}{4}.$$
Thus, the number of acute triangles is at most
$$\binom{n}{3} - n \cdot \frac{(n-2)^2}{4} \cdot \frac{1}{2} = \frac{n(n-2)(n+2)}{24}.$$
The regular polygon case does not yield equality. Indeed, we first consider a regular polygon $B_1B_2\cdots B_n$. Then we rotate half of the points, say $B_1, B_2, \dots, B_{\frac{n}{2}}$, anticlockwise about the centre of the circle with a sufficiently small angle $\theta$ to points $C_1, C_2, \dots, C_{\frac{n}{2}}$. Then $C_1C_2\cdots C_{\frac{n}{2}}B_{\frac{n}{2}+1}B_{\frac{n}{2}+2}\cdots B_n$ is one possible case which gives the upper bound $\frac{n(n-2)(n+2)}{24}$.