It seems like those classic problems for lunch, i.e., despite being easy, is very fun. Suppose there are $n$ girls and $m$ boys, say $g_1, g_2, ..., g_n$ are the $n$ girls and $b_1, b_2, ..., b_m$ are the $m$ boys. Assume the statement is incorrect, in other words, every boy $b_i$ won their match against every girl $g_j,$ therefore, every boy's loss was against another boy, then the different amounts of losses vary from $0$ to $m-1,$ however it's said everyone lost at least once, implying no boy won against all the other boys, applying Pigeon Hole Principle follows there are $2$ boys with the same amount of losses, a contradiction. Hence, the statement is correct.

Suppose otherwise that every boi won every match against every gurl. Then, each boi will have lost at most $b-1$ matches, where $b$ is the total amount of bois. Though, since each boi has lost at least one match, there are $b-1$ possible amount of matches lost by any boi and $n$ bois with distinct amounts of matches lost, which is impossible.