Inequality with condition $\sum\limits_{1\leq i,j\leq n}|1-x_ix_j|=\sum\limits_{1\leq i,j\leq n}|x_i-x_j|$ 
Let $x_1,x_2,\ldots,x_n\, (n\geq 3)$ be positive real numbers such that
$$\sum_{1\leq i,j\leq n}|1-x_ix_j|=\sum_{1\leq i,j\leq n}|x_i-x_j|.$$
Prove that: for arbitrary real number $a_1,a_2,\ldots,a_n$, there is a real number $t$, such that
$$\sum_{i=1}^n |\sin (t-a_i)|\leq \cot \left(\frac\pi{2\sum_{i=1}^n x_i}\right).$$

I think $\sum_{1\leq i,j\leq n}|1-x_ix_j|=\sum_{1\leq i,j\leq n}|x_i-x_j|$ is difficult to use, because it seems so surprising.
 A: This edition gives a second and better counterexample. 
As you say, the given condition on the $x_1,x_2,x_3,\cdots,x_n$ is difficult to use.
I'm afraid what you're proposing is not true. Here a counterexample for $n=3$. Because of the difficulty mentioned above the calculations are approximate but the continuity makes them very plausible.
Take $(x_1,x_2,x_3)\approx(0.1,\space 1.432,\space 0.02)$. You have with these values $$(|x_1-x_2|+|x_1-x_3|+|x_2-x_3|)-(|1-x_1x_2|+|1-x_1x_3|+|1-x_2x_3|)\approx 0.01$$ In other words $RHS-LHS\approx 0.01$  and with $x_1,x_2,x_3$ having better decimal approximation would have the equality.
Now one has
$$2(x_1+x_2+x_3)\approx 3.104\\\cot\left(\dfrac{\pi}{3.104}\right)\approx 0.6251195670$$ 
and taking the arbitraries $(a_1,a_2,a_3)=\left(\dfrac{\pi}{2},\dfrac{\pi}{5},\dfrac{\pi}{5}\right)$ one has the function
$$g(x)=|\sin(x-\dfrac{\pi}{2})|+2|\sin(x-\dfrac{\pi}{5})|\ge 0.831$$ For which, whatever $x$ ($t$ in your notation) we would have the absurdity
$$0.831\le 0.625$$
A better second counterexample: $(x_1,x_2,x_3)=(0.3,\space0.07,\space 1.3157)$.
Now you have $$LHS-RHS=0.000791$$ and 
$\cot\left(\dfrac{\pi}{3.3714}\right)\approx 0.7429287$ which with the same $g(x)$ above gives the absurde $$0.831\le0.7429$$
►This counterexample is better than the first one because $LHS$ is closer to $RHS$◄ 
A: Proof.
From the condition $\sum_{1\leq i,j\leq n}|1-x_ix_j|=\sum_{1\leq i,j\leq n}|x_i-x_j|$, we have
$$\sum_{i=1}^n x_i = n.$$
Proof: It is example 15 in page 443,
“Problems from the book” (2008) by Titu Andreescu and Gabriel Dospinescu. Also, see the proof by Ravi B@AoPS, by applying USAMO 2000 Problem 6.
Thus, we need to prove that, for arbitrary real numbers $a_1, a_2, \cdots, a_n$, there is a real number $t$ such that
$$\sum_{i=1}^n |\sin (t - a_i)| \le \cot \frac{\pi}{2n}.$$
WLOG, assume that $0\le a_1 \le a_2 \le \cdots \le a_n \le \pi$.
(Note: $|\sin (t - a_i)| = |\sin (t - a_i + k\pi)|$ for all $k \in \mathbb{Z}$.)
We use @cats's very nice idea
(@Martin R told me this link).
Let $a_{n + k} = \pi + a_k$, $k = 1, 2, \cdots, n-1$.
We have
\begin{align*}
 2\sum_{1\le i < j \le n} \sin (a_j - a_i)
 &= \sum_{j=1}^{n-1} \sum_{i=1}^n \sin(a_{i+j} - a_i)\\
 &\le \sum_{j=1}^{n-1} n \sin \frac{\sum\limits_{i=1}^n (a_{i+j} - a_i)}{n}  \tag{1}\\
 &= \sum_{j=1}^{n-1} n \sin \frac{\pi j}{n}\\
 &= \frac{n}{2\sin \frac{\pi}{2n}} \sum_{j=1}^{n-1} 2\sin \frac{\pi}{2n} \sin \frac{\pi j}{n}
 \\
 &= \frac{n}{2\sin \frac{\pi}{2n}} \sum_{j=1}^{n-1}
 \left(\cos\frac{(2j-1)\pi}{2n} - \cos\frac{(2j+1)\pi}{2n}\right)\\
 &= \frac{n}{2\sin \frac{\pi}{2n}}\left(\cos\frac{\pi}{2n} - \cos\frac{(2(n-1)+1)\pi}{2n}\right)\\
 &= n\cot \frac{\pi}{2n}. \tag{2}
\end{align*}
Explanation: (1)
$x\mapsto \sin x$ is concave on $[0, \pi]$.
And $a_{i+j} - a_i \in [0, \pi]$ for all $i=1, 2, \cdots, n$
and $j=1, 2, \cdots, n-1$. Apply Jensen's inequality.
Let
$$f(t) = \sum_{i=1}^n |\sin (t - a_i)|.$$
Using (2), we have
$$\sum_{i=1}^n f(a_i)
= 2 \sum_{1\le i < j \le n} \sin (a_j - a_i) \le n \cot \frac{\pi}{2n}.$$
Thus, by the Pigeonhole Principle, there exists $k$ such that
$f(a_k) \le \cot \frac{\pi}{2n}$.
We are done.
