The answer is all evens. For evens, just draw out all main diagonals. Here is a sketch of the proof that odds don't work. Let's now show that odds do not work. Label the vertices $V_1, V_2, V_3, \cdots, V_n$ in clockwise order. Consider the shortest diagonal that is drawn, say $V_1 V_{i+1}$ WLOG, where $i \le \frac{n-1}{2}.$ Notice that the triangle which contains the side $V_1V_2$ must also have a side contained in $V_1 V_{i+1}.$ An analogous thing holds for $V_iV_{i+1}.$ In this way, if the triangle which contains $V_1 V_2$ is not isosceles, then we can see that the diagonals of shortest length must pair up the vertices (e.g. $(V_1, V_{i+1}$ is one pair). This is absurd. Therefore, all diagonals of the form $V_j V_{j+i}$ are drawn, for $1 \le j \le n$ (indices modulo $n$). If $i = 2$, then it's easy to see that this fails. If $i = \frac{n-1}{2}$, it's easy to see that there is an $n-$gon in the center of the polygon, absurd. Else, suppose that $2 < i < \frac{n-1}{2}.$ Note that the diagonals $V_1 V_{i+1}$ and $V_2 V_{i+2}$ intersect at an angle of $\frac{2 \pi}{n}.$ Therefore, one angle of the triangle which "tiles" the polygon is at most $\frac{2\pi}{n}.$ By looking at the triangle containing $V_1 V_2$, this implies that $i = 3.$ But then, it's easily checked that the triangle bounded by $V_1 V_4, V_2 V_5, V_3V_6$ is smaller than that bounded by $V_1V_4, V_{n-1}V_2, V_1 V_2.$ This is absurd. $\square$

i would argue other wise

Attachments:

a diagonal is an edge connecting 2 vertices.

oops my bad, didn't read it correctly