8. The sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}=(-1)^{n} n-a_{n}, n=1,2,3, \cdots$, and $a_{10}=a_{1}$, then the maximum value of $a_{n} a_{n+1}$ is $\qquad$ .
According to the problem, $a_{n+1}+a_{n}=(-1)^{n} n$, let $n=1,2, \cdots, 9$ we get
$$
\left\{\begin{array} { l } 
{ a _ { 2 } + a _ { 1 } = - 1 , } \\
{ a _ { 3 } + a _ { 2 } = 2 , } \\
{ a _ { 4 } + a _ { 3 } = - 3 , } \\
{ a _ { 5 } + a _ { 4 } = 4 , } \\
{ \cdots \cdots } \\
{ a _ { 1 0 } + a _ { 9 } = - 9 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a_{2}+a_{1}=-1, \\
-a_{3}-a_{2}=-2, \\
a_{4}+a_{3}=-3, \\
-a_{5}-a_{4}=-4, \\
\cdots \cdots \\
a_{10}+a_{9}=-9
\end{array} \Rightarrow a_{10}+a_{1}=-\frac{9 \times 10}{2}=-45,\right.\right.
$$

Also, $a_{10}=a_{1} \Rightarrow a_{1}=-\frac{45}{2}$. Following the above steps, we get
when $n$ is odd, $-a_{n}+a_{1}=-\frac{n(n-1)}{2} \Rightarrow a_{n}=\frac{n(n-1)-45}{2}$;
when $n$ is even, $a_{n}+a_{1}=-\frac{n(n-1)}{2} \Rightarrow a_{n}=-\frac{n(n-1)-45}{2}$.
Thus, $a_{n} a_{n+1}=-\frac{1}{4}[n(n-1)-45][n(n+1)-45]=-\frac{1}{4}\left[\left(n^{2}-45\right)-n\right]\left[\left(n^{2}-45\right)+n\right]$ $=-\frac{1}{4}\left[\left(n^{2}-45\right)^{2}-n^{2}\right]=-\frac{1}{4}\left(n^{4}-91 n^{2}+45^{2}\right)$.
By the properties of quadratic functions, when $n=7$, $a_{n} a_{n+1}$ reaches its maximum value $\frac{33}{4}$.