Problem 187

Note that since set $A$ has $m$ consecutive integers that sum to $2m$, the middle integer (i.e., the median) must be $2$. Therefore, the largest element in $A$ is $2 + \frac{m-1}{2}$.
Further, we see that the median of set $B$ is $0.5$, which means that the "middle two" integers of set $B$ are $0$ and $1$. Therefore, the largest element in $B$ is $1 + \frac{2m-2}{2} = m$. $2 + \frac{m-1}{2} > m$ if $m < 3$, which is clearly not possible, thus $2 + \frac{m-1}{2} < m$.
Solving, we get \begin{align*} m - 2 - \frac{m-1}{2} &= 99\\ m-\frac{m}{2}+\frac{1}{2}&=101\\ \frac{m}{2}&=100\frac{1}{2}.\\ m &= \boxed{201}\end{align*}