Four. (50 points) A football invitational tournament has 29 cities participating, 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, a statistical review found that, except for Team A from City $A$, all other teams have played a different number of matches. How many matches has Team B from City $A$ played? Prove your conclusion.
Four, Solution: Team B from City A has played 28 matches.
First, let's discuss the problem in general. Suppose there are $n$ cities participating in the competition that meet the conditions of the problem. Let $a_{n}$ denote the number of matches Team B from City A has played. In the case of $n$ cities, according to the competition rules, each team can play at most $2(n-1)$ matches. By the problem's conditions, the number of matches played by the $2n-1$ teams other than Team A from City A should be $0, 1, 2, \cdots, 2(n-1)$. If $a_{n}=2(n-1)$, then this team has played against all other teams, so none of them would have played zero matches, which is a contradiction. Therefore, $a_{n} \neq 2(n-1)$. For the sake of discussion, let's assume Team A from City B has played $2(n-1)$ matches. This means that all other teams, except for Team A from City B, have played at least one match, so Team B from City B has played zero matches. Removing the two teams from City B, we consider the remaining $n-1$ cities. Each of these teams has played one match against Team A from City B and none against Team B from City B. Therefore, among these $n-1$ cities, the number of matches played by each team, except for Team A from City A, is: $1-1=0, 2-1=1, \cdots, 2(n-1)-1=2n-3$, and they are all distinct. Since in the case of $n-1$ cities, the number of matches played by Team B from City A is $a_{n}$, we have $a_{n}=a_{n-1}-1, n \geqslant 2$, and clearly $a_{1}=0$. Thus, the sequence: $a_{1}, a_{2}, \cdots, a_{n}$ is an arithmetic sequence with the first term $a_{1}=0$ and a common difference of 1. Therefore, $a_{n}=n-1$. When $n=29$, $a_{29}=29-1=28$. Hence, Team B from City A has played 28 matches.