Nice problem!

Let $A[boy, girl] = 1$ if boy and girl danced with each other, and $A[boy, girl] = 0$ otherwise. Let us fix some boy $B$. Let us compute in two ways the number of elements in the set: $$S = \{(g_1,g_2,b) \mid A[b,g_1] + A[b,g_2] + A[B,g_1]+A[B,g_2] \equiv 1 \pmod 2, g_1 < g_2\}.$$From our main assumption in this problem we have that : for each pair of girls $(g_1,g_2)$ we have exactly $n$ boys such $(g_1,g_2,b) \in S$. Thus $$|S| = (2n-1)n^2.$$On the other hand: $$|S| = \sum_{b=1}^{2n} |\{g \mid A[b,g]+A[B,g] \equiv 0\pmod 2\}| \quad \cdot \quad |\{g \mid A[b,g]+A[B,g] \equiv 1\pmod 2\}|$$but $|\{g \mid A[b,g]+A[B,g] \equiv 0\pmod 2\}| \quad + \quad |\{g \mid A[b,g]+A[B,g] \equiv 1\pmod 2\}| = 2n$ so from AM-GM, $$|S| = 0 + \sum_{1\leq b \leq n, b \neq B} |\{g \mid A[b,g]+A[B,g] \equiv 0\pmod 2\}| \quad \cdot \quad |\{g \mid A[b,g]+A[B,g] \equiv 1\pmod 2\}| \leq (2n-1) \cdot n^2$$Therefore in each AM-GM we have equality and thus $|\{g \mid A[b,g]+A[B,g] \equiv 0\pmod 2\}| = n$ for each $b$, which we wanted to prove.

A More Combinatorial Solution Let's create the obvious bipartite graph which connects a boy and a girl iff they danced. Let the girls be $g_1, g_2, \cdots, g_{2n}$ and the boys be $b_1, b_2, \cdots, b_{2n}$. Call a quartet consisting of two boys and two girls $odd$ if there are an odd number of edges amongst them. We claim that the condition of the problem is equivalent to the graph attaining the maximal number of $odd$ quartets, which would then solve the problem since maximizing the number of odd quartets is symmetric. Let $f(i, j)$ be the number of odd quartets containing girls $g_i, g_j$, and let $g(i, j)$ be the number of boys which danced with exactly one of $g_i, g_j$. We claim that $f(i, j) = g(i, j)(2n- g(i, j))$ in general. To see this, note that in any odd quartet containing girls $g_i, g_j, b_\alpha, b_\beta,$ we have that exactly one of $b_\alpha, b_\beta$ danced with one of $g_i, g_j$ (and the other danced with both or none of $g_i, g_j$). Furthermore, any pair $b_\alpha, b_\beta$ of boys for which exactly one of them danced with exactly one of $g_i, g_j$ also forms a odd quartet together with $g_i, g_j$. Hence, the claim that $f(i, j) = g(i, j)(2n-g(i,j))$ immediately follows. Since $\sum_{i, j} f(i, j)$ counts the number of odd quartets, it is now immediate that the number of odd quartets is maximized iff $g(i, j) = n$ for all $i, j$ (the only if part comes from the fact that $g(i, j) = n$ is possible). An aanalogous argument for boys with $b(i, j)$, etc. solves the problem.