【Solution】Solution: According to the problem:

From the relationship between the remainder and the divisor, we know that Good < Double. Let the common divisor of Good and Double be $d$.
Then we have $(\overline{\text { Good }} \div \mathrm{d}) \times 99=(\overline{\text { Double }} \div d) \times N$.
Since $\overline{\text { Good }} \div d$ and $\overline{\text { Double }} \div d$ are coprime, $\overline{\text { Double }} \div d$ must be a divisor of 99. Therefore, the result of $\overline{\text { Double }} \div d$ is
9 (11, 1, and 99, 33, and 3 do not meet the requirements).
To maximize Good and Double, $\overline{\text { Double }}=9 \times d$, Good $=8 \times \mathrm{d}$. When $d=9$, Good and Double reach their maximum value of 7281.

Therefore, the answer is: 7281
9. (10 points) A four-digit number "HaoShiChengShuang" divided by a two-digit number "ChengShuang" has a remainder exactly equal to "HaoShi". If different Chinese characters represent different digits and "HaoShi" and "ChengShuang" are not coprime, then the largest four-digit number "HaoShiChengShuang" is