9. Let the set $A=\left\{\left.\frac{a_{1}}{9}+\frac{a_{2}}{9^{2}}+\frac{a_{3}}{9^{3}}+\frac{a_{4}}{9^{4}} \right\rvert\, a_{i} \in\{0,1,2, \cdots, 8\}, i=1\right.$, $2,3,4\}$, arrange the numbers in $A$ in descending order, and find the 1997th number.
9. The elements in $A$ are of the form $\frac{1}{9^{4}}\left(a_{1} \cdot 9^{3}+a_{2} \cdot 9^{2}+a_{3} \cdot 9+a_{4}\right)$, where the number inside the parentheses is represented in base 9 as $\left(a_{1} a_{2} a_{3} a_{4}\right)_{9}$. The smallest number is 0, and the largest number is $(8888)_{9}=6560$, so $A$ contains 6561 elements. Thus, the 1997th largest number is the 6561 - 1997 + 1 = 4565th smallest number, which is the decimal number 4564, and $4564=(6231)_{9}$. Therefore, the required number is $\frac{6}{9}+\frac{2}{9^{2}}+\frac{3}{9^{3}}+\frac{1}{9^{4}}$