jmc

Note that $20m=\underset{\text{9 digits}}{\underbrace{444444440}}$, so $n=20m+4$.

If $d$ is any common divisor of $m$ and $n$, then $d$ must also be a divisor of $n-20m=4$. Therefore, $\gcd (m,n)$ is either $1$, $2$, or $4$. We can see that $m$ is not divisible by $4$ (since its last two digits form $22$, which is not a multiple of $4$). However, both $m$ and $n$ are clearly divisible by $2$, so $\gcd (m,n)=\boxed{2}$.