XVIII OBM

No, it's not possible. Let $S$ be the set of $n$ points. Consider the smallest circumcircle $\omega$ of all triangles determined by the $n$ points. Let $A,B,C\in S$ be three points on $\omega$ and let $O$ be its center. If one of the central angles $\angle AOB$, $\angle BOC$, $\angle COA$ is less than $120^{\circ }$ then the circumcircle of the correspondent triangle is smaller than $\omega$. Since the sum of the three central angles is $360^{\circ }$, $\angle AOB=\angle BOC=\angle COA=120^{\circ }$. The circumcenters of $AOB$, $BOC$ and $COA$ are the midpoints of the arcs $AB$, $BC$, $CA$, and then there are six points from $S$ on $\omega$ forming a regular hexagon. Take $O$ and two of these points and obtain a circumcircle smaller than $\omega$, which is a contradiction. So there is no such set.