8. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{1}{3}, a_{n+1}=\frac{a_{n}}{2-(2 n+1)(2 n-5) a_{n}}, n=1,2, \cdots$, then $\sum_{i=1}^{2023} a_{i}=$
$$
\begin{array}{l}
\frac{1}{a_{n+1}}=\frac{2}{a_{n}}-(2 n+1)(2 n-5)=\frac{2}{a_{n}}-4 n^{2}+8 n+5 \\
\Rightarrow \frac{1}{a_{n+1}}-\left[a(n+1)^{2}+b(n+1)+c\right]=2\left[\frac{1}{a_{n}}-\left(a n^{2}+b n+c\right)\right] \\
\Rightarrow \frac{1}{a_{n+1}}-\left[a n^{2}+(2 a+b) n+a+b+c\right]=\frac{2}{a_{n}}-\left(2 a n^{2}+2 b n+2 c\right) \\
\Rightarrow \frac{1}{a_{n+1}}=\frac{2}{a_{n}}-a n^{2}+(2 a-b) n+a+b-c \Rightarrow a=4, b=0, c=-1 .
\end{array}
$$

Then $\frac{1}{a_{n}}-\left(4 n^{2}-1\right)=\left[\frac{1}{a_{1}}-(4-1)\right] \cdot 2^{n-1}=0 \Rightarrow a_{n}=\frac{1}{4 n^{2}-1}$.
So $\sum_{i=1}^{2023} \frac{1}{4 i^{2}-1}=\frac{1}{2} \sum_{i=1}^{2023}\left(\frac{1}{2 i-1}-\frac{1}{2 i+1}\right)=\frac{1}{2}\left(1-\frac{1}{4047}\right)=\frac{2023}{4047}$.