I. solution. Let $A$ be the most successful player of the first team, that is, the one who won the most matches in the first team. (If there are more such players, let $A$ be one of them.) According to the condition, $A$ had a loss, and one of the players who defeated him is $B$. Since $B$ defeated $A$, he could only lose to a $C$ different from $A$.

Since $A$ is the best player in his team, there is a player $D$ in the second team whom $A$ defeated, and from whom $C$ lost. If there were no such player, then $C$ would have won more matches than $A$, since he also defeated $B$. This proves our statement, because the players $A, B, C, D$ defeated each other in a cycle.

II. solution. If we line up the participants of the matches starting from any player, such that everyone is followed by a player they defeated, then the line will eventually form a circle, meaning we will eventually reach a player who defeated someone who is ahead of them. As long as this does not happen, the line cannot break, since everyone won at least one match. Therefore, there are players who defeated each other in a cycle. Each such group can only consist of an even number of players, since there are an equal number of players from each team. We will show that the smallest such group consists of four players.

A group of two players is clearly not possible. If our statement were not true, there would be a number $N$ greater than 4, such that there is an $N$-player group that defeated each other in a cycle, but no group with fewer than $N$ players. Take an $N$-player group that defeated each other in a cycle, and let $A$ be any of its members, and $B, C, D$ be the players defeated by $A, B, C$ respectively. If $D$ defeated $A$, then $A, B, C, D$ would form a group that defeated each other in a cycle. If $D$ lost to $A$, we could remove $B$ and $C$ from the original $N$-player group, and the group would still be a cycle of players who defeated each other. Both cases lead to a contradiction with our assumption, so the smallest group of players who defeated each other in a cycle can only consist of four players.

Remarks. 1. In the second solution, we only used the fact that everyone won at least once.

2. The statement can also be proven by induction on the number of players in the teams.