Any solution , I think I have seen this problem many times.

We will prove by induction on $n$ to show that $n = 3k$ is true Suppose that all the value of $k < n$ satisfy is $k = 3q$. We will show that $ n $ = $3m$ Suppose that the $n$ - gon satisfy the conditions Let the $n$ - gon be $A_1A_2...A_n$ Because it is divided in to triangles so there exist $A_i$ such that $A_i$ is connected with $A_{i+2}$. Suppose that is $A_1,A_3$ Let $A_k$ be the point connected with $A_3$ such that between $A_1,A_k$ there is no point connected with $A_3$ If $A_k \equiv A_n$. Now we consider the $n-2$-gon $A_3A_4...A_n$ We denote $x_i$ is the number of diagonals at the point $A_i$ in $A_3A_4...A_n$. It's obvious that $\sum x_i $ is divisible by $2$ But $x_i$ is even for all $3 \leq i \leq n$ and $x_n$ is odd because $A_nA_3$ is a diagonal so $\sum x_i$ is odd which is impossible $A_k \neq A_n$ Because $A_3$ is not connected with any point between $A_1$ and $A_k$ so $A_1$ is connected with $A_k$. We consider $2$ polygons $P = A_3A_4...A_k$ and $Q = A_{k+1}..A_1$ If the number of diagonals start at $A_k$ to $P$ is even and so does $Q$ then $P,Q$ also satisfy the condition. Then, the number of vertices of $P$ and $Q$ is divisible by $3$ . So, $n$ is divisible by $3$ If the number of diagonal start at $A_k$ to $P$ is odd. Prove similary with the first case we will get a contradition Then $3|n$

@above: Is there any motivation for thinking of $3|n$?

I just tried for $n = 4,5,6,7,8,9$ and it's satisfy with $n = 6,9$ then i guess that $ n = 3k$ satisfy the conditión