We'll construct each of the sets $B$. Let $\Gamma(S)$ be the sum of the squares of the elements of $S$ and define. Then it suffices to construct $B$ such that \[ \Gamma(B) = \frac{1}{2} \sum_{i=1}^{48n+24} i^2 .\](We'll define the sum of the squares up to $48n+24$ as $f(n)$. Proceed by induction on $n$. For the base case of $n=0$, note that $B=\{24, 23, 22, 21, 20, 2 \}$ works. Assume that there exists a set $B_{n-1}$ such that $\Gamma(B_{n-1}) = \frac{1}{2}f(n)$. Then it suffices to find a subset $T$ of $\{48n-23, 48n-22, \dots 48n+24 \}$ such that $\Gamma(T) = \frac{1}{2} \Gamma(\{48n-23, 48n-22, \dots 48n+24 \})$ I claim \[ T = \{ 48n+k \mid k \in K \} \]works, where $K = \{2, 3, \dots 17, -17, -18, \dots -23 \} $. The proof is a simple calculation. \begin{align*} \Gamma(T) &= \sum_{k \in K} (48n+k)^2 \\ &= \sum_{k \in K} (48^2 n^2 + 96 kn + k^2 \\ &= |{K}| 48^2 n^2 + 96 n \sum_{k \in K} k + \sum_{k \in K} k^2 \\ &= 24 \cdot 48^2 n^2 + 96 n \cdot 12 + \left( 17^2 - 1^2 + \sum_{j=1}^{23} j^2 \right) \\ &= 24 \cdot 48^2 n^2 + 96 n \cdot 12 + \left( 288 + \sum_{j=1}^{23} j^2 \right) \\ &= \frac{1}{2}\left( 48 \cdot 48^2 n^2 + 24 \cdot 96n + 24^2 + 2 \sum_{j=1}^{23} j^2 \right) \\ &= \frac{1}{2}\Gamma(\{48n-23, 48n-22, \dots 48n+24 \}) \end{align*} Thus $B_{n-1} \cup T$ works as $B_n$ and the induction is complete.