9. The function $f(x)=$ $\frac{x+1}{x}+\frac{x+2}{x+1}+\cdots+\frac{x+2023}{x+2022}$ has a center of symmetry at $(a, b)$, then $2 a+b=$ $\qquad$ _.
$$
f(x)=2023+\frac{1}{x}+\frac{1}{x+1}+\cdots+\frac{1}{x+2022}=2023+g(x) \text {, }
$$

where $g(x)$ satisfies $g(-x-2022)=-g(x) \Rightarrow g(x)$ has a symmetry center at $(-1011,0)$, thus $f(x)$ has a symmetry center at $(-1011,2023)$. Therefore, $2 a+b=1$.