Hint: Take the largest then see what could you do with it.

Very interesting both condition and inequality because $\frac{5}{3}$ is the best upper bound which can't be achieved, but for greater $n$, we can get closer to that bound since $1+2\sum_{i=1}^{\infty} 4^{-i}=\frac{5}{3}$. Unfortunately, I still don't have a proof. Does anyone have?

IT is Wrong. Put a_1=2, a_2~a_n set nearby 0. Am I wrong? Tell me what is wrong at my analysis please.

It never said $i\neq j$. So for $i=j$ we get $x_ix_j \leq 4^{-0}$, i.e. $x_i \leq 1$ for all $i$.

Yesterday I remembered this problem and saw that it is still unsolved here, so let me type the outline of my solution. Let $x_i=4^a_i$. Then condition is equivalent to: $a_i+a_j \le -|i-j| \quad (1)$. Now, let's say that sequence $u_1,\dots, u_n \in\mathbb{R}$ "almost majorizes" sequence $v_1,\dots v_n \in\mathbb{R}$ if: $\sum_{i=1}^{k} u_i \ge \sum_{i=1}^{k} v_i$ for $k=1,2, \dots n$ (note that just equality criterion in majorization is weakened) Good thing about Karamata's inequality is that it is still valid for monotone increasing convex function and almost-majorizing sequences. Now we just need to observe that sequence $0,-1,-1,-2,-2,-3,-3 \dots $ almost majorizes every sequence $a_n$ satisfying $(1)$, so Karamata's inequality gives: \[ \sum_{i=1}^{n} x_i =\sum_{i=1}^{n} 4^{a_i} \leq 4^0+4^{-1}+4^{-1}+ \dots < 1+2\sum_{k=1}^{\infty} 4^{-k} =\frac{5}{3} \]

Why $\sum_{i=1}^{n} 4^{a_i} \leq 4^0+4^{-1}+4^{-1}+ \dots $?