11. (20 points) Let 2018 real numbers $a_{1}, a_{2}, \cdots, a_{2018}$ satisfy
$$
\sum_{i=1}^{2018} a_{i}=0, \sum_{i=1}^{2018} a_{i}^{2}=2018 .
$$

Find the maximum possible value of the minimum product of any two of these 2018 real numbers.
11. Let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{2018}$.

If $a_{1} \geqslant 0$, then $\sum_{i=1}^{2018} a_{i} \geqslant 0$.
Since $\sum_{i=1}^{2018} a_{i}=0$, it follows that $a_{i}=0 (i=1,2, \cdots, 2018)$.
Thus, $\sum_{i=1}^{2018} a_{i}^{2}=0 \neq 2018$.
Therefore, $a_{1} < 0$.
For $1 \leqslant i < j \leqslant 2018$, if $a_{i} < 0$, then $a_{2018} a_{1} < a_{j} a_{i}$;
if $a_{i} > 0$, regardless of the value of $a_{j}$, $a_{2018}(-a_{1}) \geqslant a_{j}(-a_{i})$, i.e., $a_{2018} a_{1} \leqslant a_{i} a_{j}$.
Therefore, among all the products, $a_{1} a_{2018}$ is the smallest. For $1 \leqslant k \leqslant 2018$, we always have
$$
\begin{array}{l}
\left(a_{2018}-a_{k}\right)\left(a_{k}-a_{1}\right) \geqslant 0 \\
\Rightarrow\left(a_{1}+a_{2018}\right) a_{k}-a_{k}^{2} \geqslant a_{1} a_{2018} .
\end{array}
$$

For $k=1,2, \cdots, 2018$, summing the above inequality gives
$$
-2018 \geqslant 2018 a_{1} a_{2018} \Rightarrow a_{1} a_{2018} \leqslant -1 \text{. }
$$

When $a_{1}=a_{2}=\cdots=a_{1009}=-1, a_{1010}=a_{1011}=\cdots=a_{2018}=1$, the equality holds.
Thus, the maximum value of $a_{1} a_{2018}$ is -1.