7. Let 2018 real numbers $a_{1}, a_{2}, \ldots, a_{2018}$ satisfy $\sum_{i=1}^{2018} a_{i}=0, \sum_{i=1}^{2018} a_{i}^{2}=2018$, the maximum possible value of the minimum product of any two of these 2018 real numbers is $\qquad$
7. -1

Analysis: Let's assume $a_{1} \leq \mathrm{L} \leq a_{t} \leq 0 < a_{t+1} \leq \mathrm{L} \leq a_{2018}$, then $\sum_{i=t+1}^{2018} a_{i} = -\sum_{i=1}^{t} a_{i}$; hence
$$
\begin{aligned}
2018 = \sum_{i=1}^{2018} a_{i}^{2} & = \sum_{i=1}^{t} a_{i}^{2} + \sum_{i=t+1}^{2018} a_{i}^{2} \\
& \leq a_{1} \sum_{i=1}^{t} a_{i} + a_{2018} \sum_{i=t+1}^{2018} a_{i} = -a_{1} \sum_{i=t+1}^{2018} a_{i} - a_{2018} \sum_{i=1}^{t} a_{i}, \\
& \leq -a_{1} a_{2018}(2018-t) - t a_{2018} a_{1} = -2018 a_{1} a_{2018}
\end{aligned}
$$

Thus, $a_{1} a_{2018} \leq -1$.

When $a_{1} = a_{2} = \mathrm{L} = a_{1009} = -1, a_{1010} = a_{1011} = \mathrm{L} = a_{2018} = 1$, the equality holds.