Proof Set $g(0) = -\sum_{i=1}^{n} x_i$, $g(k) = \sum_{i=1}^{k} x_i - \sum_{i=k+1}^{n} x_i$ ($1 \le k \le n-1$), $g(n) = \sum_{i=1}^{n} x_i$.
Then
$$|g(1) - g(0)| = 2|x_1| \le 2,$$
$$|g(k+1) - g(k)| = 2|x_{k+1}| \le 2, \quad k = 1, 2, \dots, n-2,$$
$$|g(n) - g(n-1)| = 2|x_n| \le 2.$$
So for each $0 \le k \le n-1$,
$$|g(k+1) - g(k)| \le 2. \qquad \textcircled{1}$$
If the conclusion is not true, by the condition for each $k$, $0 \le k \le n$, we have
$$|g(k)| > 1. \qquad \textcircled{2}$$
If there is an $i$, $0 \le i \le n-1$, such that $g(i)g(i+1) < 0$, we may assume that $g(i) > 0$ and $g(i+1) < 0$. By ②, $g(i) > 1$ and $g(i+1) < -1$. Thus $|g(i+1) - g(i)| > 2$. This contradicts ①. Thus $g(0)$, $g(1)$, ..., $g(n)$ have the same sign. But $g(0) + g(n) = 0$. The contradiction implies that the conclusion is true.