\[\begin{aligned}\sum_{i=1}^n\sum_{j=1}^n(n-|i-j|)x_ix_j&=\sum_{i=1}^{n}(x_1+x_2+\cdots+x_i)^2+\sum_{i=2}^n(x_i+x_{i+1}+\cdots+x_n)^2 \\&=\frac{x_1^2+x_n^2}{2}+\sum_{i=1}^{n-1}\frac{1}{2}((x_1+\cdots+x_i)^2+(x_1+\cdots+x_{i+1})^2) \\&+\sum_{i=1}^{n-1}\frac{1}{2}((x_i+\cdots+x_n)^2+(x_{i+1}+\cdots+x_n)^2) \\&\geq\frac{x_1^2+x_n^2}{2}+\sum_{i=1}^{n-1}\frac{1}{4}x_{i+1}^2+\sum_{i=1}^{n-1}\frac{1}{4}x_i^2 \\&\geq\frac{1}{2}\sum_{i=1}^nx_i^2. \end{aligned}\]letting $n=2k+1,x_1=1,x_2=-2,x_3=2,x_4=-2,\ldots,x_{2k}=-2,x_{2k+1}=1$ gives $c\leq\frac{1}{2}$. so $c_{max}=\frac{1}{2}.$ also an exercise using Linear Algebra(Rayleigh-Ritz)

The expression on the LHS seems quite similar to that in ISL 2015 A3 and hence the beginning of the above solution is common for the both problems.

The skill is very popular so it is not very difficult now, The main step is $$\sum_{1\le i<j\le n}(j-i)x_ix_j=\sum_{1\le i\le k<j\le n}x_ix_j=\sum_{k=1}^{n-1}\left(\sum_{i=1}^{k}x_i\right)\left(\sum_{j=i+1}^{n}x_j\right)$$And let $S_j=x_1+x_2+\ldots +x_j$, the rest is just Cauchy.

Let $s=\sum_{i=1}^{n}x_i$ and denote $y_k=\dfrac12\left(-\sum_{l=1}^{k}x_l+\sum_{l=k+1}^{n}x_l\right)$ with $y_0=-y_n=\dfrac s2$. \begin{align*} \sum_{i,j}(n-|i-j|)x_ix_j=&s^2+\sum_{k=1}^{n-1}\left((\sum_{l=1}^{k}x_l)^2+(\sum_{l=k+1}^{n} x_l)^2\right)\\=&s^2+\sum_{k=1}^{n-1}(2y_k^2+\dfrac{s^2}2)\\=&\dfrac{n-1}2s^2+\sum_{k=1}^{n}(y_{k-1}^2+y_k^2)\\\ge&0+\dfrac12\sum_{k=1}^{n}(y_{k-1}-y_k)^2=\dfrac12\sum_{k=1}^{n}x_k^2\end{align*}with equality case same as above.