9. Given that $a, b$ are positive integers, $a<b$, and $a, b$ are coprime. If the inequality $a x+b y \leqslant a b$ has exactly 2023 positive integer solutions for $x, y$, then $(a, b)=$. (Find all possible pairs that satisfy the condition)
Solve
$a x+b y \leqslant a b \Rightarrow \frac{x}{b}+\frac{y}{a} \leqslant 1$, as shown in the figure.
$(a, b)=1$ implies that there are no integer points on the line segment $\frac{x}{b}+\frac{y}{a}=1(x>0$, $y>0)$,
then the number of positive integer solutions to the inequality $a x+b y \leqslant a b$ is $\frac{(a-1)(b-1)}{2}$. According to the problem,
$$
\begin{array}{l}
(a-1)(b-1)=4046=2 \times 7 \times 17^{2} \\
\Rightarrow a-1=1,2,7,2 \times 7,17,2 \times 17,
\end{array}
$$

Thus $(a, b)=(2,4047),(3,2024),(8,579),(15,290),(18,239),(35,120)$, upon inspection, $(a, b)=(15,290),(35,120)$ do not meet the requirements.
Therefore, $(a, b)=(2,4047),(3,2024),(8,579),(18,239)$.