China Western Mathematical Olympiad

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 $$|g(1)-g(0)|=2|x_1|\le 2,$$ $$|g(k+1)-g(k)|=2|x_{k+1}|\le 2,k=1,2,\dots ,n-2,$$ $$|g(n)-g(n-1)|=2|x_n|\le 2.$$ So for each $0\le k\le n-1$, $$|g(k+1)-g(k)|\le 2.\stackrel{◯}{1}$$ If the conclusion is not true, by the condition for each $k$, $0\le k\le n$, we have $$|g(k)|>1.\stackrel{◯}{2}$$ 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.