Given $x_1, \cdots, x_n \in (0, 1)$ such that $\sum_{i \ne j}\frac{1}{x_i - x_j} = \frac{1}{1 - x_i} - \frac{1}{x_i}$, calculate $\sum_{i = 1}^nx_i$. 
Given positive integer $n$ and $x_1, x_2, \cdots, x_{n - 1}, x_n \in (0, 1)$ such that for all $i = \overline{1, n}$, we always have that $$\large \sum_{i \ne j}\frac{1}{x_i - x_j} = \frac{1}{1 - x_i} - \frac{1}{x_i}$$. Calculate the value of $\displaystyle \sum_{i = 1}^nx_i$.

We have that
$$\sum_{j \ne 1}\frac{1}{x_1 - x_j} = \frac{1}{1 - x_1} - \frac{1}{x_1}$$
$$\sum_{j \ne 2}\frac{1}{x_2 - x_j} = \frac{1}{1 - x_2} - \frac{1}{x_2}$$
$$\vdots$$
$$\sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor - 1}\frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor - 1} - x_j} = \frac{1}{1 - x_{\left\lfloor\frac{n}{2}\right\rfloor - 1}} - \frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor - 1}}$$
$$\sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor}\frac{1}{x_\left\lfloor\frac{n}{2}\right\rfloor - x_j} = \frac{1}{1 - x_\left\lfloor\frac{n}{2}\right\rfloor} - \frac{1}{x_\left\lfloor\frac{n}{2}\right\rfloor}$$
$$\implies \sum_{j \ne 1}\frac{1}{x_1 - x_j} + \sum_{j \ne 2}\frac{1}{x_2 - x_j} + \cdots + \sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor - 1}\frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor - 1} - x_j} + \sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor}\frac{1}{x_\left\lfloor\frac{n}{2}\right\rfloor - x_j}$$
$$ = \sum_{i = 1}^\left\lfloor\frac{n}{2}\right\rfloor\dfrac{1}{1 - x_i} - \sum_{i = 1}^\left\lfloor\frac{n}{2}\right\rfloor\dfrac{1}{x_i}$$
Similarly, $$\implies \sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor + 1}\frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor + 1} - x_j} + \sum_{j \ne \left\lfloor\frac{n}{2}\right\rfloor + 2}\frac{1}{x_{\left\lfloor\frac{n}{2}\right\rfloor + 2} - x_j} + \cdots + \sum_{j \ne n - 1}\frac{1}{x_{n - 1} - x_j} + \sum_{j \ne n}\frac{1}{x_n - x_j}$$
$$ = \sum_{i = \left\lfloor\frac{n}{2}\right\rfloor + 1}^n\dfrac{1}{1 - x_i} - \sum_{i = \left\lfloor\frac{n}{2}\right\rfloor + 1}^n\dfrac{1}{x_i}$$
But that's all I got.
 A: Let $P(x) = \prod_{k=1}^n (x-x_k) = x^n - Ax^{n-1} + \cdots$ and by
Vieta's formula, $\displaystyle\;A = \sum_{i=1}^n x_i$
In terms of $P(x)$, the horrible sum on LHS equals to
$$\sum_{j=1,\ne i}^n \frac{1}{x_i-x_j} = \frac12 \frac{P''(x_i)}{P'(x_i)}$$
This implies following rational function and hence its numerator vanishes at $n$ points $x_1,\ldots,x_n$.
$$\frac12\frac{P''(x)}{P'(x)} - \frac{1}{1-x} + \frac{1}{x} = 
\frac{(4x-2)P'(x) + (x^2-x)P''(x)}{2x(x-1)P'(x)}
$$
The numerator is a polynomial of degree at most $n$. So it is a multiple of $P(x)$. Comparing coefficients of $x^n$, we obtain
$$(4x-2)P'(x) + (x^2-x)P''(x) - n(n+3)P(x) = 0$$
Throwing the expansion $P(x) = x^n - Ax^{n-1} + \cdots$ into above expression and extract the coefficient of $x^{n-1}$, we obtain
$$(2n+2)A - (n^2+n) = 0\quad\implies\quad \sum_{i=1}^n x_i = A = \frac{n}{2}$$
Update
For a solution without calculus, multiply both sides by $x_i(1-x_i)$ and sum over $i$. One obtain:
$$\sum_{i,j;j\ne i}\frac{x_i - x_i^2}{x_i - x_j} = \sum_{i=1}^n(2 x_i - 1) = 2A - n$$
Swapping the summation index $i,j$ in LHS and take averages, we obtain
$$\begin{align}
{\rm LHS} = \sum_{i,j;j\ne i}\frac{x_i - x_i^2}{x_i - x_j}
&= \frac12\sum_{i,j;j \ne i}\frac{(x_i - x_i^2) - (x_j - x_j^2)}{x_i - x_j}\\
&= \frac12\sum_{i,j;j \ne i}(1 - x_i - x_j)\\
&= \frac12\left[\sum_{1 \le i,j \le n}(1 - x_i - x_j) - \sum_{i=1}^n (1 - 2x_i)\right]\\
&= \frac12\left[(n^2 - 2nA) - (n - 2A)\right]
\end{align}
$$
Combine with RHS $= 2A - n$, we get
$$n^2 - 2nA - n + 2A = 4A - 2n
\iff n(n+1) = 2(n+1) A$$
This leads to $A = \frac{n}{2}$ again.
