Edit. Eventhough I had in mind the version of the ACM I needed, it all messed up when I tried to find a reference. As for the valid version, I will refer to Vasile Cirtoaje's Algebraic Inequalities. Old and new Methods, p. $267$.
We will employ a powerful technique developed by Vasile Cirtoaje back in $2006$ called the Arithmetic Compensation Method (see, for instance, this document).
Let to this end $$F(x_1, \ldots, x_n):=\sum_{1\leqslant i<j\leqslant n}\frac{x_ix_j}{(1-x_i)(1-x_j)}$$ which is clearly symmetric and continuous on $S:=\left\{(x_1, \ldots,x_n)\mid \sum_{i=1}^n x_i=\frac12, \forall i: x_i\geqslant 0\right\}$. We will now refer to the Remark 1.1 from the document linked above, which basically states that
If $$F(x_1,x_2,x_3,\ldots,x_n)>F\left(\frac{x_1+x_2}2, \frac{x_1+x_2}{2}, x_3,\ldots,x_n\right)\label{(i)}\tag{i}$$
implies $F(x_1,x_2,x_3,\ldots,x_n)\leqslant F(0, x_1+x_2, x_3, \ldots, x_n)$, then $$F(x_1,x_2,x_3,\ldots,x_n)\leqslant \underset{1\leqslant k\leqslant n}\max F\left(\frac1{2k}, \ldots, \frac1{2k},0,\ldots ,0\right)$$
Notice that (\ref{(i)}) is equivalent to (where we will denote, for brevity, $y:=\frac{1}2(x_1+x_2)$)
\begin{align*}
\frac{x_1x_2}{(1-x_1)(1-x_2)}-\frac{y^2}{(1-y)^2}+\left(\frac{x_1}{1-x_1}+\frac{x_2}{1-x_2}-\frac{2y}{1-y}\right)\sum_{i=3}^n \frac{x_i}{1-x_i}>0\\
\iff \frac{y^2-x_1x_2}{(1-x_1)(1-x_2)(1-y)^2}\left[2y-1+2(1-y)\sum_{i=3}^n \frac{x_i}{1-x_i}\right]>0\\
\iff 2y-1+2(1-y)\sum_{i=3}^n \frac{x_i}{1-x_i}>0
\end{align*}
Where the last equivalence follows from $y^2-x_1x_2\geqslant 0$. We will now turn to the second inequality:
\begin{align*}
F(x_1,x_2,x_3,\ldots,x_n)- F(0, x_1+x_2, x_3, \ldots, x_n)\leqslant0\\
\iff \frac{x_1x_2}{(1-x_1)(1-x_2)}+\left(\frac{x_1}{1-x_1}+\frac{x_2}{1-x_2}-\frac{2y}{1-y}\right)\sum_{i=3}^n \frac{x_i}{1-x_i}\leqslant0\\
\iff \frac{x_1x_2}{(1-x_1)(1-x_2)(1-2y)}\left[1-2y+2(y-1)\sum_{i=3}^n \frac{x_i}{1-x_i}\right]\leqslant 0\\
\end{align*}
Which is easily seen to follow from (\ref{(i)}). Thus, we conclude that \begin{align*}
F(x_1,x_2,x_3,\ldots,x_n)&\leqslant \underset{1\leqslant k\leqslant n}\max F\left(\frac1{2k}, \ldots, \frac1{2k},0,\ldots ,0\right)\\
&= \underset{1\leqslant k\leqslant n}\max \binom{k}{2}\frac{1}{(2k-1)^2}\\
&= \underset{1\leqslant k\leqslant n}\max \frac{k(k-1)}{2(2k-1)^2}
\end{align*}
But $f: x\mapsto \frac{x(x-1)}{2(2x-1)^2}$ is increasing on $[1,\infty)$, and, hence, the result follows.
Remark. Based on the difficulty of the other contest problems, this solution seems a bit overkill to me, but I have failed in the attempt to find a simpler method, since most well-known inequalities work in the other direction...
Best Answer
OP's work is correct however $x_j>0$ needs to be mentioned. Here is another approach:
Let $s=\sum_{j=1}^{n} x_j,$ then $$f(x_j)=\frac{x_j}{s-x_j} \implies f''(x)=\frac{2s}{(s-x_j)^3}>0,\quad 0<x_j<s.$$ So by Jensen's inequality $$\frac{1}{n}\sum_{j=1}^{n} f(x_j) \ge f(\frac{s}{n}) \implies \sum_{j=1}^{n} \frac{x_j}{s-x_j}\ge n \frac{s/n}{s-s/n}=\frac{n}{n-1}. $$