1. Given real numbers $x_{1}, x_{2}, \cdots, x_{2018}$ satisfy: for any $1 \leqslant i<j \leqslant 2018$, we have $x_{i}+x_{j} \geqslant(-1)^{i+j}$. Find the minimum value of $\sum_{i=1}^{2018} i x_{i}$.
(Provided by Zehui Shi)
1. The minimum value sought is $\frac{2037171}{2}$.

On one hand, when $x_{1}=x_{2}=\cdots=x_{2018}=\frac{1}{2}$, the condition is satisfied, and we have
$$
\sum_{i=1}^{2018} i x_{i}=\frac{1009 \times 2019}{2}=\frac{2037171}{2} .
$$

On the other hand, we prove by mathematical induction: For any $n(n \geqslant 7)$, we have
$$
\sum_{i=1}^{n} i x_{i} \geqslant \frac{n(n+1)}{4} .
$$

When $n=7$, from the condition we know
$$
\begin{array}{l}
2\left(x_{2}+x_{6}\right) \geqslant 2,4\left(x_{4}+x_{6}\right) \geqslant 4, \\
x_{1}+x_{7} \geqslant 1, x_{3}+x_{5} \geqslant 1, \\
2\left(x_{3}+x_{7}\right) \geqslant 2,4\left(x_{5}+x_{7}\right) \geqslant 4 .
\end{array}
$$

Adding these inequalities, the conclusion holds.
Assume the conclusion holds for $n(n \geqslant 7)$, consider the case for $n+1$.

Let $y_{i}=x_{i+1}(i=1,2, \cdots, n)$. Then for any $1 \leqslant i<j \leqslant n$, we have
$$
\begin{array}{l}
y_{i}+y_{j}=x_{i+1}+x_{j+1} \\
\geqslant(-1)^{(i+1)+(j+1)}=(-1)^{i+j} .
\end{array}
$$

By the induction hypothesis,
$$
\sum_{i=1}^{n} i y_{i} \geqslant \frac{n(n+1)}{4} \text {, }
$$

which means $\sum_{i=2}^{n+1}(i-1) x_{i} \geqslant \frac{n(n+1)}{4}$.
From the given condition,
$$
x_{i}+x_{i+2} \geqslant 1(i=1,2, \cdots, n+1),
$$

with the convention $x_{n+2}=x_{1}, x_{n+3}=x_{2}$.
Adding these $n+1$ inequalities, we get
$$
2 \sum_{i=1}^{n+1} x_{i} \geqslant n+1 \Rightarrow \sum_{i=1}^{n+1} x_{i} \geqslant \frac{n+1}{2} \text {. }
$$
(1) + (2) gives
$$
\sum_{i=1}^{n+1} i x_{i} \geqslant \frac{n(n+1)}{4}+\frac{n+1}{2}=\frac{(n+1)(n+2)}{4} \text {. }
$$

By mathematical induction, the general conclusion is proved.