Let $V_1, V_2, V_3 \cdots V_n$ be the vertices of the polygon. WLOG $V_1V_2 = V_2V_3 = 1$ then by triangle inequality on $V_1V_2V_3$, $V_1V_3 < V_1V_2 + V_2V_3$ If $n\ge 4$ then $V_1V_3 = 1$ then $V_1V_2V_3$ is an equilateral triangle. Suppose that $n \ge 6$ then $V_1V_4$ and $V_2V_4$ are diagonals then triangle inequality on triangle $V_1V_2V_4$ implies $V_1V_4 = V_2V_4$ Similarly $V_1V_5 = V_2V_5$ however $V_3$, $V_4$ and $V_5$ are on the perpendicular bisector of $V_1V_2$ Contradiction, therefore $n \le 5$ and similarly with $V_2V_5 = V_3V_5$ to get the example for $n = 5$ and concave polygon if you add the convex condition $n = 4$ is the maximum.