Choose $n$ out of $2n-1$ boxes containing at least half of all white balls and half of all black balls
Let $B^*, B_1, B_2, \dots, B_{2n-2}$ be the the bins sorted by the number of white balls, in non-increasing order (break ties arbitrarily). Notice that the first bin is called $B^*$.

Consider the two groups $G_E = \{ B^* \} \cup \{B_i \mid i \text{ is even} \}$ and $G_O = \{ B^* \} \cup \{B_i \mid i \text{ is odd} \}$.
Clearly $|G_E| = |G_O| = 1 + \frac{2n-2}{2} = n$.

Moreover, each of $G_E$ and $G_O$ contains at least $W/2$ white balls.
This is true since since for each bin not in $G_E$ (resp. $G_O$) the previous bin in the sorted order must be in $G_E$ (resp. $G_O$). That is, the total number of white balls in $G_E$ (resp. $G_O$) is at least as large as the number of white balls not in $G_E$ (resp. $G_O$).

Now, total the number of black balls in at least one group $G^* \in \{G_E, G_O\}$  must be at least $B/2$. This is easy to see since $\{G_E, G_O \setminus \{B^*\}\}$
is a partition of the set of bins, and hence the bins in at least one set of the partition (which is either $G_E$ or a subset of $G_O$) must contain at least half the black balls.