Solution:

The $m$ integers of $\mathcal{A}$ have mean $\frac{2m}{m} = 2,$ so they are centered at $2;$ since the mean of consecutive integers is an integer only when there are an odd number of them, $m$ is odd and the greatest element of $\mathcal{A}$ is $2 + \frac{m-1}{2}.$ The $2m$ integers of $\mathcal{B}$ have mean $\frac{1}{2},$ so they are $1 - m, \ldots, 0, 1, \ldots, m,$ with greatest element $m.$

The condition is $$\left| 2 + \frac{m-1}{2} - m \right| = \left| \frac{3 - m}{2} \right| = 99,$$ so $|3 - m| = 198,$ giving $m = 201$ (since $m \gt 0$). Indeed $201$ is odd, as required, so $m = 201.$