19 In a carriage, any $m(m \geqslant 3)$ passengers have a unique common friend (if A is a friend of B, then B is also a friend of A, and no one is a friend of themselves), how many friends does the person with the most friends have in this carriage?


19 Let the person with the most friends have $k$ friends, clearly $k \geqslant m$. If $k>m$, let $A$ have $k$ friends $B_{1}$, $B_{2}, \cdots, B_{k}$. And denote
$$
S=\left\{B_{1}, B_{2}, B_{3}, \cdots, B_{k}\right\},
$$

Let $B_{i_{1}}, B_{i_{2}}, \cdots, B_{i_{m-1}}$ be any $m-1$ elements chosen from $S$. Then $A, B_{i_{1}}, \cdots, B_{i_{m-1}}$ these $m$ people have a common friend, denoted as $C_{i}$. Since $C_{i}$ is a friend of $A$, it follows that $C_{i} \in S$. 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 $\left\{A, B_{i_{1}}, \cdots, B_{i_{m-1}}\right\}$ and $\left\{A, B_{j_{1}}, \cdots, B_{j_{m-1}}\right\}$ are $C_{i}$, $C_{j} \in S$, respectively, then it must be that $C_{i} \neq C_{j}$, 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 $C_{i}$, which contradicts the problem statement. In this way, under the above correspondence, each element in $S$ corresponds to at most one $(m-1)$-element subset of $S$, thus we have
$$
C_{k}^{m-1} \leqslant k
$$

Since $m \geqslant 3, m-1 \geqslant 2$, we have
$$
C_{k}^{m-1}>C_{k}^{1}=k
$$

Clearly, (1) and (2) are contradictory. Therefore, the maximum value sought is $m$.