Proof by contradiction. Assume that any two three-person committees either have two members in common or no members in common.

If committees $A$ and $B$ have members in common, then they have two common members $a$ and $b$. If committee $B$ and $C$ also have (two) common members, then at least one of $a, b$ belongs to $C$, thus $C$ and $A$ also have (two) common members. Therefore, committees with common members can be grouped together, such that within the same group, each committee has two common members, and committees from different groups have no common members.

The number of committees in each group $k$ must not exceed the number of members in that group $h$. In fact, it is clear that $h \geqslant 3$. When $h=3$, $k=1 \leqslant h$; when $h \geqslant 4$, $k \geqslant 2$.

Suppose $\{x, y, a\},\{x, y, b\}$ are two committees in the same group, then the other committees in this group can only be of the form $\{x, a, b\},\{y, a, b\}$, or $\{x, y, d\}$, where $d$ has at most $h-4$ choices, so $k \leqslant 4+(h-4)=h$. Thus, the total number of committees $n+1 \leqslant$ the number of people $n$, which is a contradiction.