Final Round

Answer: 7 sets. Let $N$ be the required number of 3-element sets satisfying the problem condition. Suppose that $N\ge 8$. Let $M=\{x,y,z\}$ be one of these sets. Since the number of the remaining sets is greater than or equal to 7 and each such set has exactly one common element with $M$, we see that there exists an element of $M$ (say, $x$) that belongs to at least $[7/3]=3$ sets. Let the sets $M_1,M_2,M_3$ contain $x$. By condition, there is not an element that belongs to all sets, so there exists a set (say, $M_0$) such that there is an element from $M$ (say, $y$), which belongs to $M_0$. It follows that $x\notin M_0$. Since any two of the sets $M_1,M_2,M_3$ and $M$ have not common elements except for $x$ (and $x\notin M_0$), so all four elements of the intersection of $M_0$ with $M_1,M_2,M_3$, $M$ are distinct. Therefore $M_0$ consists of at least four elements, a contradiction. Thus, $N\le 7$.

It remains to show the example for $N=7$. The required sets (see the Fig.) are $\{A,K,B\}$, $\{A,P,M\}$, $\{A,N,C\}$, $\{B,M,C\}$, $\{B,P,N\}$, $\{C,P,K\}$, $\{K,M,N\}$.