Let there be $x$ white and $y$ black balls. Since $\frac{10}{11}x$ is an integer, $x$ is divisible by $11$. Next, the $x - \frac{10}{11}x = \frac{x}{11}$ balls not in a mixed pair in the first division must be paired up among themselves. Hence $\frac{x}{11}$ is even, i.e. $x$ is even. Thus $x$ is divisible by $22$. Similar observations on the second division show that $y$ is divisible by $26$.

In order to pair up $\frac{10}{11}x$ white balls with black ones it is necessary to have at least $\frac{10}{11}x$ black balls, i.e. $\frac{10}{11}x \le y$. Likewise $\frac{12}{13}y \le x$, or $y \le \frac{13}{12}x$. In summary $22|x$, $26|y$ and $\frac{10}{11}x \le y \le \frac{13}{12}x$. Conversely, if $x$ and $y$ satisfy these conditions then both divisions are possible.

The multiples of $22$ in $[150, 200]$ are $154$, $176$ and $198$. Since $22|x$ and $150 \le x \le 200$, we have $x \in \{154, 176, 198\}$. For $x = 154$, $x = 176$, $x = 198$ the condition $\frac{10}{11}x \le y \le \frac{13}{12}x$ yields $y \in [140, 166]$, $y \in [160, 190]$, $y \in [180, 214]$ respectively. Taking $26|y$ into account we obtain 4 solutions: $(154, 156)$, $(176, 182)$, $(198, 182)$, $(198, 208)$.