$\begin{array}{r}\text { I. (40 points) Given } x_{i} \in \mathbf{R}(1 \leqslant i \leqslant 2020), \\ \text { and } x_{1010}=1 \text {. Find the minimum value of } \sum_{i, j=1}^{2020} \min \{i, j\} x_{i} x_{j} \text { . }\end{array}$
$$
\begin{array}{l} 
\sum_{i, j=1}^{2020} \min \{i, j\} x_{i} x_{j} \\
= \sum_{i=1}^{2020} i x_{i}^{2}+2 \sum_{1 \leq i<j \leqslant 2020}^{2020} i x_{i} x_{j} \\
=\left(x_{1}+x_{2}+\cdots+x_{2020}\right)^{2}+ \\
\left(x_{2}+x_{3}+\cdots+x_{2020}\right)^{2}+\cdots+ \\
\left(x_{2019}+x_{2020}\right)^{2}+x_{2020}^{2} .
\end{array}
$$

Let $\sum_{i=1011}^{2020} x_{i}=x$. Then
$$
\begin{array}{l}
\sum_{i, j=1}^{2020} \min \{i, j\} x_{i} x_{j} \geqslant(x+1)^{2}+x^{2} \\
=2 x^{2}+2 x+1 \geqslant \frac{1}{2},
\end{array}
$$

When $x_{1009}=\frac{1}{2}, x_{1011}=-\frac{1}{2}, x_{1010}=1$, and the rest $x_{i}=0$ $(i \neq 1009,1011)$, the equality in (1) holds.