11. (20 points) Let $a_{i}\left(i \in \mathbf{Z}_{+}, i \leqslant 2020\right)$ be non-negative real numbers, and $\sum_{i=1}^{2020} a_{i}=1$. Find the maximum value of $\sum_{\substack{i \neq j \\ i l j}} a_{i} a_{j}$.
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 $\sum_{\substack{i \neq j \\ i \mid j}} a_{i} a_{j}$ 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 $\sum_{i \in A} a_{i}=1$, and when $A=\{1,2,4, \cdots, 1024\}$, $|A|_{\text {max }}=11$.
$$
\begin{aligned}
& \text { By }\left(\sum_{i \in A} a_{i}\right)^{2}=\sum_{i \in A} a_{i}^{2}+2 \sum_{i<j} a_{i} a_{j} \\
& \geqslant \frac{1}{|A|}\left(\sum_{i \in A} a_{i}\right)^{2}+2 \sum_{i<j} a_{i} a_{j}, \\
& \text { hence } \sum_{\substack{i \neq j \\
i \mid j}} a_{i} a_{j} \leqslant \sum_{\substack{i \neq j \\
i, j \in A}} a_{i} a_{j} \\
\leqslant & \frac{1}{2}\left(1-\frac{1}{|A|}\right)\left(\sum_{i \in A} a_{i}\right)^{2} \\
= & \frac{1}{2}\left(1-\frac{1}{|A|}\right) \\
\leqslant & \frac{5}{11} .
\end{aligned}
$$