jmc

The highest that $FLYFLY$ can be would have to be $124124$, and it cannot be higher than that because then it would exceed the $6$-digit limit set on $BUGBUG$. So, if we start at $124124\cdot 8$, we get $992992$, which would be wrong because both $B\&U$ would be $9$, and the numbers cannot be repeated between different letters. If we move on to the next highest, $123123$, and multiply by $8$, we get $984984$. All the digits are different, so $FLY+BUG$ would be $123+984$, which is $1107$. So, the answer is $\boxed{1107}$.