This solution is due to Michael Kural and David Stoner. Let $F = \sum_{1 \le r < s \le 2n} (s-r-n) x_rx_s$ be the objective function. We claim $F \le n(n-1)$ is the maximum, achieved when $x_i = (-1)^i$. By linearity, we can assume $x_i \in \{\pm1\}$. First, define the partial sums \begin{align*} a_i &= x_1 + \dots + x_i \\ b_i &= x_{2n} + \dots + x_i \end{align*}for each $i=1,\dots,2n$ (thus each $a_i$ and $b_i$ is an integer). Also, let $c = x_1 + \dots + x_{2n} = a_{2n} = b_1$. Then we have \begin{align*} \sum_{i=1}^{2n} (a_i^2 + b_i^2) &= \sum_{1 \le r < s \le 2n} \left( (2n+1-s) + r \right)2x_rx_s + \sum_{1 \le r \le 2n} (2n+1) x_r^2 \\ &= 2(n+1) \sum_{1 \le r < s \le 2n} x_rx_s - 2F + (2n+1) \sum_{1 \le r \le 2n} x_r^2 \\ &= (n+1) c^2 - 2F + n \sum_{1 \le r \le 2n} x_r^2 \\ &= (n+1) c^2 - 2F + 2n^2. \end{align*}Thus \[ 2F \le (n+1)c^2 + 2n^2 - \sum_{i=1}^{2n} (a_i^2+b_i^2) \]Now, by computing the difference $a_i^2 + b_{2n+1-i}^2 - \frac{1}{2} c^2$ for $i=1,2,\dots,2n-1$, (which is of the form $X^2+Y^2-\frac{1}{2} (X+Y)^2 = \frac{1}{2} (X-Y)^2$), we find that \begin{align*} \sum_{i=1}^{2n} (a_i^2+b_i^2) - (n+1)c^2 &= \frac{1}{2}(x_1-x_2-x_3-\dots-x_{2n-1}-x_{2n})^2 \\ &+ \frac{1}{2}(x_1+x_2-x_3-\dots-x_{2n-1}-x_{2n})^2 \\ &+ \frac{1}{2}(x_1+x_2+x_3-\dots-x_{2n-1}-x_{2n})^2 \\ &+ \dots \\ &+ \frac{1}{2}(x_1+x_2+x_3+\dots+x_{2n-1}+x_{2n})^2. \end{align*}Within each consecutive pair of these sums, one of the sums is at least $2$, contributing $\frac{1}{2} \cdot 2^2 = 2$. Thus we conclude that the entire sum must be at least $2n$. So, $2F \le 2n^2 - 2n \implies F \le n(n-1)$ as desired.

Here is the Inspiring solution given by the shortlist (First three lines from solution posted above) Let $F =\sum_{1 \le r < s \le 2n} (s-r-n) x_rx_s$ be the objective function. We claim $F \le n(n-1)$ is the maximum, achieved when $x_i = (-1)^i$. By linearity, we can assume $x_i \in \{\pm1\}$. First, define the partial sums \begin{align*} a_i &= x_1 + \dots + x_i \\ b_i &= x_{2n} + \dots + x_i \end{align*}for each $i=1,\dots,2n$ (thus each $a_i$ and $b_i$ is an integer). Now for all integers $1 \le k \le 2n$ define $T_k = (a_{i-1}-b_{i+1})^2$ (here $a_0 = 0$ and $b_{2n+1} = 0$) Consider $Q=\sum_{p=1}^{2n} T_p$ In the full expansion of this sum notice that for $1 \le r < s \le 2n, x_rx_s$ the coefficient is $$2(r-1)+2(2n-s)-2(s-r-1) = -4s+4n+4r = - 4(s-r-n)$$ Thus $Q = -4F + (2n-1) \sum_{i=1}^{2n} (x_i)^2 = -4F + (2n-1)(2n) = -4F+4n^2-2n$ and so $4F = 4n^2-2n-Q$. Thus, we wish to minimize $Q$ Now remark that each $T_k$ is $(\text{odd number of terms which are -1 or 1})^2$ meaning we have $T_k \ge 1 \implies Q \ge 2n$ But this means $4F \le 4n^2-2n-2n = 4n^2-4n$ or $F \le n^2-n = n(n-1)$ as desired. oops thanks @below

@above wrong inequality signs on the last line- should be $\le$ not $\ge$

Darn. That's a horrible looking expression. Call that $F$ like the above. Of course, looking at this expression as a linear function of $x_i$ for each $x_i$, we can assume $x_i = \pm 1$. Okay so that $x_rx_s$ term implies that we have to square some stuff up. Take the partial sum $s_i = \sum_{k=1}^i x_k$. Then $\sum_{i=1}^{2n} s^2_i$ will have $2(2n+1-s)x_rx_s$. So for us to have $2s-2r-2n$, we have to find another sum of squares of partial sum which gives us $2rx_rx_s$. Of course, this is the partial sum $t_i = \sum_{k=i}^{2n} x_k$. Then $\sum_{i=1}^{2n} t^2_i$ will have $2rx_rx_s$. Define $t_{2n+1}=0$. This is for later calculations. Now let's calculate this. $$\sum_{i=1}^{2n} (s^2_i+t^2_i) = (2n+1) \sum_{i=1}^{2n} x^2_i + \sum_{1\le r<s \le 2n} (4n+2-2s+2r)x_rx_s$$$$= (2n+1) \sum_{i=1}^{2n} x^2_i - 2F + (2n+2) \sum_{1\le r < s \le 2n} x_rx_s = (n+1) (\sum_{i=1}^{2n} x_i)^2 - 2F + n\sum_{i=1}^{2n} x^2_i = (n+1) (\sum_{i=1}^{2n} x_i)^2 - 2F + 2n^2$$ In another words, $$2F = 2n^2 + (n-1) (\sum_{i=1}^{2n} x_i)^2 - \sum_{i=1}^{2n-1} (s^2_i + t^2_{i+1})$$Now note that $s_i+t_{i+1} = \sum_{i=1}^{2n} x_i$. Therefore, $\frac{1}{2}(\sum_{i=1}^{2n} x_i)^2 - s^2_i - t^2_{i+1} = -\frac{1}{2}(s_i-t_{i+1})^2$. Now we have $$2F = 2n^2 - \sum_{i=1}^{2n} \frac{1}{2}(s_i-t_{i+1})^2$$Notice that $(s_i-t_{i+1})-(s_{i-1}-t_i) = 2x_i$, so $s_i-t_{i+1}$ changes by $\pm 2$ as $i$ increases by $1$. Also note that $s_i-t_{i+1} \equiv 0 \pmod{2}$. To minimize the sum of squares, it is best to have $\pm 2$, $0$, $\pm 2$, $0$ $\cdots$. This gives us $2F \le 2n^2-2n$, or $F \le n(n-1)$. To show this is possible, take $x_i$ to be $1, -1, 1, -1, \cdots$. Then $s_i=1, t_{i+1}=-1$ if $i \equiv 1 \pmod{2}$, and $s_i=t_{i+1}=0$ if $i \equiv 0 \pmod{2}$. $\blacksquare$

Hmm,isn't it hard for A3?I think A4 is easier than this? (Does anyone think like me?)

No, the main problem is figuring out convexity. After that, it is just bash.

As the expression is linear in each variable, the maximum occurs at a corner of the hypercube, i.e. when all of the $x_i$ are $\pm 1.$ We claim that the maximum is $n(n-1),$ achieved by $x_i=(-1)^i.$ The main idea is to let $\varepsilon_{i,j}=\begin{cases}1,\ |i-j|>n \\ -1,\ |i-j|<n\end{cases},$ and to write \[\sum_{1\leq r<s\leq 2n}(s-r-n)x_rx_s=\sum_{i=1}^{2n}\sum_{i+1\leq j<k\leq i+n}\varepsilon_{jk}x_{j}x_{k}.\]We can think of each term $\sum_{i+1\leq j<k\leq i+n}\varepsilon_{jk}x_jx_k.$ as an $n$-gon with consecutive vertices in the $2n$-gon shown below. With this geometric interpretation, we see that $\varepsilon_{ij}=1$ iff the segment connecting $i,j$ intersects the positive $x$-axis. Thus, taking each $x_j$ with $1<j\leq N$ to $-x_j,$ all of the coefficients become $-1.$ This is because exactly one of the endpoints of the crossing edges are switched, while $0$ or $2$ endpoints of non-crossing edges are switched. [asy][asy] int N = 10; pair[] A = new pair[N]; size(5cm); for (int i = 0; i <N; ++i) { A[i] = dir(180/N+360/N*i); } for (int i = 0; i < N; ++i) { draw(A[i]--A[(i+1)%N],lightblue); label("$"+string(i+1)+"$",A[i],dir(360/N*i+180/N)); } draw((0,0)--(1.1,0),red); draw(A[1]--A[N-1],orange+dashed); [/asy][/asy] Now $\sum_{i+1\leq j<k\leq i+n}-x_jx_k=\tfrac{x_{i+1}^2+\cdots+x_{i+n}^2-(x_{i+1}+\cdots+x_{i+n})^2}{2}=\tfrac{n-S^2}{2}.$ If $n$ is odd, $|S|\geq 1$ so each term is bounded above by $\tfrac{n-1}{2}$ and summing over all $2n$ polygons gives us the bound. If $n$ is even, analyzing the equality cases shows that the average value of $S^2$ for each term is at least $1,$ which gives us the desired bound.

The answer is $n^2-n$, which can be achieved by $(1,-1,1,-1,\ldots,1,-1)$. Since the expression we want to maximize is linear with respect to each variable, we can assume that $x_i$ is either $1$ or $-1$ for all $i=1,\ldots,2n$. The key step is to notice that $$\sum_{1 \leq r<s \leq 2n} (s-r-n)x_rx_s=(x_1)(x_2+\cdots+x_{2n})+(x_1+x_2)(x_3+\cdots+x_{2n})+\cdots+(x_1+\cdots+x_{2n-1})(x_{2n})-n\sum_{1 \leq i<j \leq 2n} x_ix_j$$Suppose that out of $x_1,x_2,\ldots,x_n$, $k$ of them are $1$ and $2n-k$ of them are $-1$. Then, $$\sum_{1 \leq i<j \leq 2n} x_ix_j=\dbinom{k}{2}+\dbinom{2n-k}{2}-k(2n-k)=2k^2-4nk+2n^2-n=2(k-n)^2-n.$$By AM-GM, we also know that $$(x_1+\cdots+x_i)(x_{i+1}+\cdots+x_{2n}) \leq \left(\frac{x_1+\cdots+x_n}{2}\right)^2=(k-n)^2.$$Now, we do casework on the parity of $k-n$. Case 1: $k-n$ is even. Since $(x_1+\cdots+x_i)(x_{i+1}+\cdots+x_{2n})$ is odd for all odd $i$ from $1$ to $2n-1$, we know that $$(x_1+\cdots+x_i)(x_{i+1}+\cdots+x_{2n}) \leq (k-n)^2-1$$for all odd $i$ in that range. Thus, we want to maximize $$n\left((k-n)^2-1\right)+(n-1)(k-n)^2-n\left(2(k-n)^2-n\right)=n^2-n-(k-n)^2.$$So, the expression is bounded above by $n^2-n$. Case 2: $k-n$ is odd. Since $(x_1+\cdots+x_i)(x_{i+1}+\cdots+x_{2n})$ is even for all even $i$ from $1$ to $2n-1$, we know that $$(x_1+\cdots+x_i)(x_{i+1}+\cdots+x_{2n}) \leq (k-n)^2-1$$for all even $i$ in that range. Thus, we want to maximize $$(n-1)\left((k-n)^2-1\right)+n(k-n)^2-n\left(2(k-n)^2-n\right)=n^2-n+1-(k-n)^2.$$Since $k-n$ is odd, the expression is bounded above by $n^2-n$, completing our proof.

What is "linearity" thing that everybody is using(or as @G_U puts it, the corner of the hypercube)? I don't understand how you can just ignore the rest of elements? Although it feels intuitive, is there like a rigorous way to show that this assumption is valid? Also what hypercube is @G_U talking about?

JustKeepRunning wrote: What is "linearity" thing that everybody is using(or as @G_U puts it, the corner of the hypercube)? I don't understand how you can just ignore the rest of elements? Although it feels intuitive, is there like a rigorous way to show that this assumption is valid? Also what hypercube is @G_U talking about? Well,for me ,I would like to write the formula as a function of x_i, and the degree is 1, which implies whenever the other elements is, the maximum occurs at the border

The motivation sort of felt like "get a vibe of what the equality cases are and then combi interp/algmanip the expression to work well with those equality cases."