67th Czech and Slovak Mathematical Olympiad

Consider a 6-good club and denote some seven of its members by $A,\dots ,G$. It suffices to show that $A,\dots ,G$ can be seated around a table as required. Consider only friendships among $A,\dots ,G$. First, we show that every member has at least three friends.

Without loss of generality consider $G$. By assumption, $B,\dots ,G$ can be seated as required, hence $G$ has at least two friends. Without loss of generality, $F$ is one of them. By assumption, $A,\dots ,E,G$ (omitting $F$) can be seated as required, hence $G$ has at least two more friends apart from $F$ for a total of at least three friends.

Since every member has at least three friends, there exists a member with at least four friends (otherwise the number of friendly pairs equals $\frac{1}{2}\cdot 7\cdot 3$, which is clearly impossible). Without loss of generality, assume $G$ has at least four friends.

By assumption, $A,\dots ,F$ can be seated as required. In such a seating, some two of the four friends of $G$ are neighbors and we can seat $G$ in between them.