Define a graph, where there is a directed edge $A \rightarrow B$, when $A$ won 2 times with $B$, and undirected edge $A - B$, when $A$ won one time and $B$ also won one time. Note that for every pair of vertices there exist exactly one (un)directed edge. Now let's deleted directed cycles, i.e. for a directed cycle $A_1 \rightarrow A_2 \rightarrow \ldots \rightarrow A_m \rightarrow A_1$, we cancel one of the matches $A_i \rightarrow A_{i+1}$, where $A_{n+1} = A_1$. Then we can ignore the cycle, and just delete it. Repeat this process. We can see that it terminates and also at the end there cannot be any directed edges. Suppose not, let $A_1\rightarrow A_2 \rightarrow \ldots \rightarrow A_k$ be the longest directed path, by the assumption we know that there exists $A_r$ such that $A_k \rightarrow A_r$, because $A_k$ lost to $A_{k-1}$. We know that $r\neq 1,2,\ldots,k-2$, because it would form a cycle, so $A_1\rightarrow A_2\rightarrow\ldots\rightarrow A_k \rightarrow A_r$ is a longer that the longest, contradiction. So we left with an undirected graph with even degree of every vertex. So for every component there exist a Euler cycle, so for every Euler cycle $B_1 - B_2 - \ldots - B_s - B_1$, cancel matches $B_i - B_{i+1}$, where $B_{s+1} = B_1$. In that way we canceled exactly have matches.