Obviously $n>3$. If $n = 2k,k \geqslant 2$, we can take regular $2k$-gon. If $n = 2k + 3,k \geqslant 2$, we can take regular $2k$-gon plus $3$ points. For $n=5$, there are $n-2=3$ triangles, so they must pairwise congruence, but almost $2$ of them contains the circumcenter (or on the sides of triangles), contadiction. So $n=4$ and $n>5$ are all solutions.

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