34th Hellenic Mathematical Olympiad

Since in each round play exactly 3 players, the number of couples playing at each round is $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$. Therefore, at the end of the game after $n$ rounds, the total number of couples played the game will be $3n$. According to the last rule: $$\left(\genfrac{}{}{0ex}{}{n}{2}\right)\le 3n\iff \frac{n(n-1)}{2}\le 3n\iff \frac{n-1}{2}\le 3\iff n\le 7.$$ Next we will prove that the value $n=7$ is possible. In fact, for $n=7$ we have $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=\frac{7!}{2!5!}=\frac{6\cdot 7}{2}=21=3\cdot 7$. If the friends are: $A$, $B$, $\Gamma$, $\Delta$, $E$, $Z$, $H$, then we can define the following triads $(A,B,\Gamma )$, $(A,\Delta ,E)$, $(A,Z,H)$, $(B,\Delta ,H)$, $(B,E,Z)$, $(\Gamma ,\Delta ,Z)$, $(\Gamma ,E,H)$.