9. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1, a_{2}=2, a_{n+2}-3 a_{n+1}+2 a_{n}=1$. Let $T_{n}$ be the sum of the first $n$ terms of the sequence $\left\{\frac{8}{a_{n}}\right\}$, then $\left[T_{2023}\right]=$ $\qquad$ . (The symbol $[x]$ denotes the greatest integer less than or equal to $x$)
Solving
$a_{3}=6-2+1=5$, since $\left\{\begin{array}{l}a_{n+2}-3 a_{n+1}+2 a_{n}=1, \\ a_{n+3}-3 a_{n+2}+2 a_{n+1}=1\end{array}\right.$
$$
\Rightarrow\left(a_{n+3}-a_{n+2}\right)-3\left(a_{n+2}-a_{n+1}\right)+2\left(a_{n+1}-a_{n}\right)=0 \text {. }
$$

Let $b_{n}=a_{n+1}-a_{n}$, then $b_{n+2}-3 b_{n+1}+2 b_{n}=0$, the characteristic equation is $x^{2}-3 x+2=0$, thus $b_{n}=c_{1}+c_{2} \cdot 2^{n}$, substituting $b_{1}=1, b_{2}=3$ gives $b_{n}=2^{n}-1$.
Therefore, $a_{n}=a_{1}+\sum_{i=1}^{n-1}\left(a_{i+1}-a_{i}\right)=1+\sum_{i=1}^{n-1}\left(2^{i}-1\right)=2^{n}-n\left(n \in \mathbf{N}^{*}\right)$.
Noting that $\frac{8}{a_{1}}+\frac{8}{a_{2}}+\frac{8}{a_{3}}+\frac{8}{a_{4}}+\frac{8}{a_{5}}=8+4+\frac{8}{5}+\frac{8}{12}+\frac{8}{27}=14+\frac{76}{135}$,
$a_{n+1}-2 a_{n}=2^{n+1}-(n+1)-2\left(2^{n}-n\right)=n-1 \Rightarrow a_{n+1} \geqslant 2 a_{n} \Rightarrow \frac{1}{a_{n+1}} \leqslant \frac{1}{2} \cdot \frac{1}{a_{n}}$
$\Rightarrow \frac{8}{a_{6}} \leqslant \frac{1}{2} \cdot \frac{8}{a_{5}}, \frac{8}{a_{7}} \leqslant \frac{1}{2} \cdot \frac{8}{a_{6}} \leqslant\left(\frac{1}{2}\right)^{2} \cdot \frac{8}{a_{5}}, \frac{8}{a_{8}} \leqslant\left(\frac{1}{2}\right)^{3} \cdot \frac{8}{a_{5}}, \cdots$,
thus $\sum_{i=6}^{2023} \frac{8}{a_{i}} \leqslant \frac{8}{a_{5}} \sum_{i=1}^{2018}\left(\frac{1}{2}\right)^{i}<\frac{8}{a_{5}}=\frac{8}{27}=\frac{40}{135}$.
Therefore, $14+\frac{76}{135}<T_{2023}=\sum_{i=1}^{2023} \frac{8}{a_{i}}<14+\frac{116}{135} \Rightarrow\left[T_{2023}\right]=14$.