In a wagon, every $m \geq 3$ people have exactly one common friend. (When $A$ is $B$'s friend, $B$ is also $A$'s friend. No one was considered as his own friend.) Find the number of friends of the person who has the most friends.
1. **Understanding the Problem:**
   We are given a graph where every set of \( m \geq 3 \) people has exactly one common friend. We need to find the number of friends of the person who has the most friends.

2. **Applying the Friendship Theorem:**
   The Friendship Theorem states that if every pair of people in a group has exactly one common friend, then there is one person who is friends with everyone else. This theorem can be extended to the case where every set of \( m \) people has exactly one common friend.

3. **Graph Structure:**
   According to the problem, the graph has a specific structure:
   - There are \( m-1 \) vertices \( x_1, x_2, \ldots, x_{m-1} \), each of which is connected to all other vertices.
   - Every other vertex has exactly one neighbor among \( x_1, x_2, \ldots, x_{m-1} \).

4. **Induction Hypothesis:**
   We use induction on \( m \). Assume the statement is true for all values smaller than \( m \) (for \( m \geq 3 \)).

5. **Base Case:**
   For \( m = 3 \), the graph must have a structure where there are 2 vertices \( x_1 \) and \( x_2 \) connected to all other vertices, and every other vertex has exactly one neighbor among \( x_1 \) and \( x_2 \).

6. **Inductive Step:**
   - Let \( x_1 \) be one of the vertices with maximal degree.
   - Apply the induction hypothesis for \( m-1 \) to the graph induced on the set of neighbors of \( x_1 \).
   - We find neighbors \( x_2, \ldots, x_{m-1} \) of \( x_1 \), each of which is connected to all other neighbors of \( x_1 \).
   - Because of the maximality assumption, the neighbors of \( x_i \) are the neighbors of \( x_1 \) for all \( i = 1, 2, \ldots, m-1 \).

7. **Common Neighbor Argument:**
   - Let \( x \) be a neighbor of \( x_i \) different from all the \( x_i \).
   - Since \( y, x_1, \ldots, x_{m-1} \) have exactly one common neighbor, \( x \) is connected to exactly one other neighbor of \( x_i \).
   - The graph \( G \) induced on the set formed by \( x_i \) and their neighbors has the form described above.

8. **Contradiction Argument:**
   - Suppose we can find a vertex \( y \) not in \( G \).
   - There must be at least one neighbor \( x \) of \( x_i \) different from all \( x_i \); otherwise, \( y, x_i \) can have no common friend.
   - Take such a neighbor \( x \). The vertices \( x_1, \ldots, x_{m-2}, x, y \) have exactly one common friend, and this friend can only be the unique neighbor of \( x_i \) linked to \( x \) (call it \( \bar{x} \)).
   - Repeat the procedure with \( \bar{x} \) instead of \( x \). We find that \( y \) is linked to \( x \) as well.
   - But now \( y, x_i \) have at least two common friends (\( x \) and \( \bar{x} \)), which is a contradiction.

9. **Conclusion:**
   This means that \( x_i \) and their neighbors are all the vertices in the graph, and we have what we wanted.

The final answer is \( \boxed{ m-1 } \).