Example 2 In a football invitational tournament, 16 cities participate, each city sending Team A and Team B. According to the competition rules, each pair of teams plays at most one match, and the two teams from the same city do not play against each other. After several days of the tournament, it was found that except for Team A of City $A$, all other teams have played a different number of matches. How many matches has Team B of City $A$ played? Please prove your conclusion.
$(1985$, National High School Mathematics Competition)
Solution: Consider the general case: Suppose there are $n$ cities participating, and let the number of matches played by Team B from City A be $a_{n}$. Clearly, $a_{1}=$ 0.

Since there are $n$ cities participating, according to the competition rules, each team can play at most $2(n-1)$ matches. Given the conditions, the number of matches played by the $2 n-1$ teams, excluding Team A from City A, covers $0,1,2, \cdots, 2(n-1)$. Assume Team A from City B has played $2(n-1)$ matches, meaning that except for Team B from City B, all other teams have played at least one match. Therefore, the number of matches played by Team B from City B is 0.

Now, remove the two teams from City B, and for the remaining $n-1$ cities, the number of matches played by the teams, excluding Team A from City A, are respectively:
$$
\begin{array}{l}
1-1=0, \\
2-1=1, \\
\cdots \cdots, \\
(2 n-3)-1=2(n-2) .
\end{array}
$$

The number of matches played by Team B from City A is $a_{n}-1$,
thus $a_{n-1}=a_{n}-1(n \geqslant 2)$.
It is evident that the sequence $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term 0 and common difference 1, hence,
$$
a_{n}=0+(n-1) \cdot 1=n-1 \text {. }
$$

In this problem, $n=16$, so, $a_{16}=15$, which means Team B from City A has played 15 matches.