We have the summation
$$ Z = \sum_{1 \le r < s \le 2n}^{2n} f(r,s) x_{r} x_{s} $$
The possibilities of $x_{i}$s are bounded by $-1 \le x_{i} \le 1$. Also, $f(r,s)$ real function. How to prove the argument: "since $Z$ is linear in the variable $x_{i}$, then the maximum occurs at the end points. ($x_{i} \in \{-1,1\}$)"?
My analysis is that when all $f(r,s)$s are positive, then of course maximum is when $x_{r}x_{s} =1$. But $f(r,s)$ are not all positive.
It is actually quite easy to show. Let the maximum be attained for the collection of values $x_1,\dots,x_{2n}$. Let us look at a particular term $x_r$ for some $1 \leq r \leq 2n$ and the summands that contain it. Consider the sum $S = x_1 x_r + x_2 x_r + \dots + x_{r-1}x_r + x_{r+1}x_r + \dots + x_{2n} x_r = (x_1 + x_2 + \dots + x_{r-1} + x_{r+1} + \dots + x_{2n})x_r = Tx_r$.
If we vary $x_r$ between $-1$ and $1$, the terms in the sum $Z$ that are not in the sum $S$ do not change. So in order to maximize $Z$, we need to find value of $x_r$ that maximizes $S$. Now, if $T = 0$, the maximum of $S$ is definitely attained at $x_r = 1$ or $x_r = -1$ (all the same). On the other hand, if $T > 0$, the maximum of is T is attained for the maximum positive value of $x_r$, which is $1$. Similarly, if $T < 0$, the maximum of $S$ is attained for negative value of $x_r$ with maximum absolute value, which is $-1$.
That is, for any collection of values $x_1, \dots, x_{2n}$, and any particular $x_r$, we can get at least as large a value of $Z$ by 'pushing $x_r$ to one-end.' That is, the maximum value of $Z$ is attained for $x_i \in \{-1,1\}$.