jmc

From $3m+4n=100$ we can get that $n=25-\frac{3}{4}m$. Then we want to minimize $$\left|m-n\right|=\left|m-25+\frac{3}{4}m\right|=\left|\frac{7}{4}m-25\right|=\left|7m-100\right|$$In other words we want $7m$ as close to $100$ as possible while still giving us integer solutions for $m$ and $n$ to the equation $3m+4n=100$.

By trial and error, we can find that the solution to $3m+4n=100$ that makes $m$ closest to $\frac{100}{7}$ is $(m,n)=(16,13)$. Then we have $\left|m-n\right|=16-13=\boxed{3}$.