HMMT November, 2022

Solution:
The median of both sets, which we will call $a$ and $b$ respectively, are more than exactly 12 of the members in their own set. Since $a$ and $b$ are consecutive, they must also be higher than the lower half of the other set and lower than the higher half of the other set, meaning that they are both higher than exactly 24 numbers in $\{1,2, \ldots, 50\}-\{a, b\}$. Thus, $a=25$ and $b=26$.

The 24 lower numbers can be divided into the two groups (with 12 in each group) in $\binom{24}{12}$ ways. Similarly, the 24 higher numbers can be divided into the two groups in $\binom{24}{12}$ ways. Thus, the answer is $\binom{24}{12}^{2}$.