ILL 1986 Problem 25

Let real numbers $x_1, x_2, \cdots , x_n$ satisfy $0 < x_1 < x_2 < \cdots< x_n < 1$ and set $x_0 = 0, x_{n+1} = 1$. Suppose that these numbers satisfy the following system of equations: $$\sum_{j=0, j \neq i}^{n+1} \frac{1}{x_i-x_j}=0 \text{where } i = 1, 2, . . ., n.$$

Prove that $x_{n+1-i} = 1- x_i$ for $i = 1, 2, . . . , n.$