National Olympiad of Argentina

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: $(154,156)$, $(176,182)$, $(198,182)$, $(198,208)$.