9. (20 points) A four-digit number $\overline{a b c d}$ is called perfect if $a+b=c+d$. How many perfect numbers can be represented as the sum of two four-digit palindromes?
Solution: Let the number $\overline{a b c d}=\overline{n m m n}+\overline{x y y x}$, then

$$
\overline{a b c d}=1001(n+x)+110(m+y) \vdots 11
$$

From the divisibility rule by 11, it follows that $b+d=a+c$. Since the number $\overline{a b c d}$ is perfect, we get $a=d$ and $b=c$, hence the original number is a palindrome, and the first digit is not equal to 1 (otherwise it cannot be represented as the sum of two four-digit numbers). All four-digit palindromes are determined by the first two digits, where the first digit can be chosen in 8 ways, and the second in 10 ways. In total, we get that there are $8 \cdot 10=80$ such numbers. It remains to note that each obtained number can be represented in the form $\overline{\overline{b b b a}}=1001+\overline{(a-1) b b(a-1)}$.

Answers: 80.