8. Let $a_{i}\left(i \in \mathbf{N}_{+}, i \leqslant 2009\right)$ be non-negative real numbers, and $\sum_{i=1}^{2009} a_{i}=1$. Then the maximum value of $\sum_{i \neq j, i l j} a_{i} a_{j}$ is
8. $\frac{5}{11}$.

If $a_{i}$ and $a_{j} (i < j, i \nmid j)$ are both not 0, then adjust $\left(a_{i}, a_{j}\right)$ to $\left(a_{i}+a_{j}, 0\right)$ or $\left(0, a_{i}+a_{j}\right)$ (this is determined by the sum of the numbers whose indices have a divisibility relationship with $a_{i}$ or $a_{j}$), at this point, the corresponding value $\sum_{i=j, i a_{j}} a_{i} a_{j}$ will not decrease. Continue such operations until the indices of the non-zero numbers all have a divisibility relationship.

Let the set of all these numbers be $A$. Then $\sum_{i \in 1} a_{i}=1$, and when $A=\left\{2^{k} \mid k=0,1, \cdots, 10\right\}$, $|A|_{\text {max }}=11$. By $\left(\sum_{i \in 1} a_{i}\right)^{2}=\sum_{i \in \Lambda} a_{i}^{2}+2 \sum_{i<j} a_{i} a_{j}$ $\geqslant \frac{1}{|A|}\left(\sum_{i \in A} a_{i}\right)^{2}+2 \sum_{i<j} a_{i} a_{j}$, then
$$
\begin{array}{l}
\sum_{i * j, 1, j} a_{i} a_{j} \leqslant \sum_{i \neq j, i, j \in A} a_{i} a_{j} \\
\leqslant \frac{1}{2}\left(1-\frac{1}{|A|}\right)\left(\sum_{i \in A} a_{i}\right)^{2} \\
=\frac{1}{2}\left(1-\frac{1}{|A|}\right) \leqslant \frac{5}{11} .
\end{array}
$$

When $a_{i}=\frac{1}{11}(i \in A), a_{j}=0(j \notin A)$, the equality holds in the above expression.