For $d=0$ any $n$ obviously works. In the sequel, assume $d\geq 1$. Consider the graph $F$ of vertices $B$ ($|B|=b$ boys) and $G$ ($|G|=g$ girls), $b+g=n$. Counting the $B-G$ edges, we get $bd = gd$, hence $b=g=n/2$, forcing $n\geq 2d$ to be even. Clearly, such a structure can be realized for any $n\geq 2d$; indexing $B=\{b_0,b_1,\ldots,b_{n/2-1}\}$ and $G=\{g_0,g_1,\ldots,g_{n/2-1}\}$ in $\mathbb{Z}_{n/2}$ allows creating edges $\{b_ig_i, b_ig_{i+1}, \ldots, b_ig_{i+d-1}\}$ for all $1\leq i \leq n/2-1$. Now, the graphs $B$ and $G$, each of order $n/2$, need each be $d$-regular, and it is folklore this is possible if and only if $d\leq n/2-1$ and $d(n/2)$ even. Therefore the "good" pairs $(n,d)$ are those with $n$ even, $n/2 \geq d+1$, and $d(n/2)$ even.