Connect two teams only if thay tied the match. Consider the adjacency matrix $M$ of this graph. The statement just says that the sum of any subset of rows is not $0$ over $\mathbb{F}_2.$ This is not possible when $n$ is odd. Indeed, $M$ has $0$'s on its main diagonal and is symmetric. You may look on $M$ also as an antisymetric matrix by putting $-1$ instead of the $1$'s that are above the main diagonal (it's the same over $\mathbb{F}_2$). The determinant of a matrix like that is $0$, when $n$ is odd because $\det(M)=\det(M^T)=\det(-M)=(-1)^n\det(M)$, so $\det(M)=0.$ This means the rows are not independent as vectors in $\mathbb{F}_2^n,$ hence there is a subset of rows that sum up to $0$ (over $\mathbb{F}_2$).