jmc

If $m\equiv 6(\mathrm{mod}9)$, then we can write $m$ as $9a+6$ for some integer $a$. This is equal to $3(3a+2)$, so $m$ is certainly divisible by $3$. If $n\equiv 0(\mathrm{mod}9)$, then $n$ is divisible by $9$. Therefore, $mn$ must be divisible by $3\cdot 9=27$.

Note that $m$ can be 6 and $n$ can be 9, which gives us $mn=54$. Also, $m$ can be 15 and $n$ can be 9, which gives us $mn=135$. The gcd of 54 and 135 is 27.

Therefore, the largest integer that $mn$ must be divisible by is $\boxed{27}$.