9.1. Let's call a four-digit number $\overline{a b c d}$ curious if the sum of the two-digit numbers $\overline{a b}$ and $\overline{c d}$ equals the two-digit number $\overline{b c}$. For example, the number 1978 is curious because 19+78=97. Find the number of curious numbers. Answer. 36.
Solution. Let's form the equation $\overline{a b}+\overline{c d}=10(a+c)+b+d=\overline{b c}=10 b+c$, from which we get $10 a+9 c+d=9 b$. The difference $9(b-c)$ is divisible by 9, so the sum $10 a+d=9(b-c)$ is also divisible by 9, which is equivalent to the divisibility by 9 of the sum $a+d=9(b-c-a)$. The sum of two different numbers from the interval from 0 to 9 is no less than 1 and no more than 17, so the latter is only possible when $a+d=9$. Substituting this into the previous equation, we get $b=a+c+1$, so the quadruple $(a, b, c, d)$ can be written in the form $(a, a+c+1, c, 9-a)$, where $a, c$ are any pair of digits satisfying the relations $a \geq 1, a+c \leq 8$. For each fixed $a=1,2, \ldots, 8$, the value of $c$ can be any from 0 to $8-a$, a total of $9-a$ options. Therefore, the total number of curious numbers is the sum $8+7+\ldots+1=36$.

Evaluation criteria. . ($\cdot$) It is proven that $a+d=9(b-c-a)$ is divisible by 9: 2 points. . ($\cdot$) It is proven that from this follows $a+d=9$: another 2 points. . ($\cdot$) The form of the quadruple of digits is written as $(a, a+c+1, c, 9-a)$: another 1 point. . ($\cdot$) The number of curious quadruples is found: 2 points.