Solution
= Solution
Let the bipartition be $X\sqcup Y$. Counting edges from each side gives
$$
k|X|=|E(G)|=k|Y|,
$$
so $|X|=|Y|=n/2$. For $S\subseteq X$, the $k|S|$ edges leaving $S$ all end in $N(S)$, while each vertex of $N(S)$ receives at most $k$ of them. Therefore
$$
k|S|\leq k|N(S)|,
$$
and Hall's condition holds. There is a matching covering $X$, of size $n/2$. No matching can use more than one edge at each vertex, so this is maximal and
$$
\boxed{\nu(G)=\frac n2.}
$$