What's answer ? Any solution or idea?

Solution. (based on the author's idea) We can assume the points $A_1,A_2,\ldots,A_n$ are ordered counterclockwise on the circle, indexed in $\mathbb{Z}_n$ (thus $A_{mn+k}=A_k$). Denote by $A_iA_j$ the arc of the circle starting from $A_i$ and ending in $A_j$ (in counterclockwise direction). We call an arc {\it non-acute} if $m(A_iA_j)\geq 180^\circ$. Let $x_s$ be the number of non-acute arcs $A_iA_{i+s}$ (having $s-1$ points in the interior). Clearly, $m(A_iA_j)+m(A_jA_i)=360^\circ$, hence at least one of the arcs $A_iA_j$ and $A_jA_i$ is non-acute, and thus we can infer that \[(*)\qquad\qquad x_s+x_{n-s}\geq n\] for every $s=1,2,\ldots,n-1$. Equality holds if and only if there are no diametrically opposite points $A_i,A_{i+s}$. We will count the number of non-acute triangles $A_iA_jA_k$. Clearly, such a triangle has exactly one angle corresponding to a non-acute arc. For each non-acute arc $A_iA_j$ having $s-1$ points in the interior there are $n-s-1$ non-acute triangles $A_iA_jA_k$, for which $\angle A_k $ corresponds to the arc $A_{i}A_j$, namely those with $A_k$ in the interior of the arc $A_jA_i$. It follows the number of non-acute triangles is \[ N=x_1\cdot (n-2)+x_2\cdot (n-3)+\cdots + x_{n-2}\cdot 1+x_{n-1}\cdot 0.\] By regrouping terms and using (*) we obtain for $n$ odd \[N\geq 0\cdot(x_{n-1}+x_1)+1\cdot(x_{n-2}+x_2)+\cdots +\dfrac{n-3}{2}\cdot(x_{\frac{n+1}{2}}+x_{\frac{n-1}{2}})\geq\] \[\geq n\left(1+2+\cdots +\dfrac{n-3}2\right)=\dfrac{n(n-3)(n-1)}8.\] Thus the number of acute triangles is at most \[\dbinom {n}{3} - \dfrac{n(n-3)(n-1)}8 = \dfrac{n(n-1)(n+1)}{24}\] and it can be reached, for example, for a regular $n$-gon. By regrouping terms and using (*) we obtain for $n$ even \[N\geq 0\cdot(x_{n-1}+x_1)+1\cdot(x_{n-2}+x_2)+\cdots +\dfrac{n-4}{2}\cdot(x_{\frac{n+2}{2}}+x_{\frac{n-2}{2}})+\dfrac{n-2}{2}\cdot x_{\frac{n}{2}}\geq\] \[\geq n\left(1+2+\cdots +\dfrac{n-4}2 +\dfrac{n-2}4 \right)=\dfrac{n(n-2)^2}8.\] Thus the number of acute triangles is at most \[\dbinom {n}{3} - \dfrac{n(n-2)^2}8 = \dfrac{n(n-2)(n+2)}{24}\] and it can be reached, for example, for an almost regular $n$-gon (with no diametrically opposite vertices).

Very nice solution ! Thank you very much mavropnevma.

Another approach. The idea is first to determine the conditions the extremal configuration should satisfy and then to calculate the number of acute triangles it contains. Assume $A_1,\ldots, A_n$ is an extremal configuration with maximal number of acute triangles. Let's denote $M=\{A_1,A_2,\ldots,A_n\}$ and for $A\in M $, $A'$ be the antipodal point of $A$. Then $A'\not\in M$, since otherwise we can slightly move $A'$ increasing the number of the acute triangles. We are going to prove that the number of points in each of the two semicircles formed by $A, A'$ are almost the same. More precisely, if $AB$ denotes the arc of the circle starting from $A$ and ending in $B$ in counterclockwise direction, we have: \[ |\,\#\{X\mid X\in M, X\in AA'\}- \#\{X\mid X\in M, X\in A'A\}\,|\le 1 \] for every $A\in M$. Assume on the contrary it's not true and let $\#\{X\mid X\in M, X\in A'A\}> \#\{X\mid X\in M, X\in AA'\}+1$. Let $B_1 \in M$ be the next to $A'$ (counterclockwise) and $B\in M$ next to $A$. Suppose $m(A'B_1) < m(AB) $ Then if we take $B_1$ and place it slightly next to $A'$ in clockwise direction, we will get a new configuration with more acute triangles, contradiction. Consider now the case $m(A'B_1) \ge m(AB) $. But then $\#\{X\mid X\in M, X\in B'B\}> \#\{X\mid X\in M, X\in BB'\}+1$ will also hold, and proceeding, we eventually will get a contradiction. It remains a routine calculation of the number of the acute triangles.