24.3 . 12 In a carriage, any $m(m \geqslant 3)$ people have a unique common friend (friendship is mutual), how many friends does the person with the most friends have? (One does not count oneself as a friend)
Let the person with the most friends have $k$ friends, obviously $k \geqslant m$. If $k > m$, let $A$ have $k$ friends $B_{1}, B_{2}, \cdots, B_{k}$. Define $S=\left\{B_{1}, B_{2}, \cdots, B_{k}\right\}$. Let $B_{i_{1}}, B_{i_{2}}, \cdots, B_{i_{m-1}}$ be any $m-1$ elements in $S$, then $A, B_{i_{1}}, B_{i_{2}}, \cdots, B_{i_{m-1}}$ these $m$ people have a common friend, denoted as $C_{i}$, then $C_{i} \in S$. If $\left\{B_{i_{1}}, B_{i_{2}}, \cdots, B_{i_{m-1}}\right\} \neq\left\{B_{j_{1}}, B_{j_{2}}, \cdots, B_{j_{m-1}}\right\}$, and the unique friends corresponding to $\Lambda, B_{i_{1}}, \cdots, B_{i_{m-1}}$ and $\Lambda$, $B_{j_{1}}, B_{j_{2}}, \cdots, B_{j_{m-1}}$ are $C_{i}$ and $C_{j}$ respectively, then $C_{i} \neq C_{j}$ must hold, otherwise $\left\{B_{i_{1}}, B_{i_{2}}, \cdots, B_{i_{m-1}}\right\} \cup\left\{B_{j_{1}}, B_{j_{2}}, \cdots, B_{j_{m-1}}\right\}$ would have at least $m$ elements, and they would have at least two friends $A$ and $C_{i}$, leading to a contradiction! In this way, under the above correspondence, each element in $S$ corresponds to at most one $(m-1)$-element subset of $S$, thus $\mathrm{C}_{k}^{m-1} \leqslant k$. However, $m \geqslant 3$, so $\mathrm{C}_{k}^{m-1}>\mathrm{C}_{k}^{1}=k$, leading to a contradiction! Therefore, the person with the most friends has $m$ friends.