How about n = 6?

Since clearly any regular polygon with $n = 3N$ sides satisfies, I bet the requirement was to show $3 \mid n$ (not $3=n$).

There was a typo from where I copied the problem - it should have been $\angle P_iP_kP_j$ rather than $\angle P_iP_jP_k$. In this case the problem seems correct. Anyway, it's a nice problem. I've only started it. Take $P_iP_j$ to be the length minimal over the set of the values of $|P_iP_j|$ - of course we secretly we know they must be all equal, so in the case that there is more than one pair $P_iP_j$ minimising this distance, choose arbitrarily. Then by the problem statement there is some $k$ such that $\angle P_iP_kP_j=60^{\circ}$. Let $P$ be a point such that $P$ is on the same half-plane formed by the line $P_iP_j$ as $P_k$ and $\triangle P_iPP_j$ is equilateral. Then $P_k$ lies on $(PP_iP_j)$. Assuming $P_k\not\equiv P_i,P_j,P$ then $P_i$ must be on one of the closed arcs $P_iP$ or $P_jP$. In the former case, $P_iP_k<P_iP=P_iP_j$ and in the latter $P_jP_k<P_jP=P_iP_j$ but both contradict the minimality of $P_iP_j$. This forces $P_k$ to be coincident with $P$ i.e. $\triangle P_iP_jP_k$ is equilateral. Now let $P_u,P_v$ be a pair of points maximising $P_uP_v$. We continue similarly as before, letting $Q$ be the point such that $\triangle P_uP_vQ$ is equilateral, and $Q$ is on the same side of $P_uP_v$ as the guaranteed vertex $P_w$ such that $\angle P_uP_wP_v=60^{\circ}$; then $Q$ cannot lie on the closed arc $P_uQ$ since otherwise $P_uP_v<P_wP_v$ which is impossible by our assumption. Repeat for the other arc to see $Q\equiv P_w$ i.e. $\triangle P_vP_wP_u$ is equilateral. We could finish the problem by proving the two triangles $P_iP_jP_k$ and $P_vP_wP_u$ are one and the same. However I've not made much progress. P.S. Is this in the right forum? Sometimes there really should be a combinatorial geometry section Edit: ok it's actually quite easy to finish off the problem. it's simple casework, using the fact that $P_1P_2\ldots P_n$ is convex.

very nice started! Finish??