Dear V.V., this Stanasila's problem is a amusement:
$ S\equiv \sum\limits_{i=1}^n \sum\limits_{j=1}^n ij\cos(x_1-x_j )=\sum\limits_{i=1}^n \sum\limits_{j=1}^n ij(\cos x_i\cos x_j +\sin x_i\sin x_j)= $
$ \sum\limits_{i=1}^n i\cos x_i\cdot\sum\limits_{j=1}^n j\cos x_j+\sum\limits_{i=1}^n i\sin x_i\cdot \sum\limits_{j=1}^n j\sin x_j=\left( \sum\limits_{i=1}^n i\cos x_i \right)^2+ $ $ \left( \sum\limits_{i=1}^n i\sin x_i \right)^2\ge 0. $
Remark. Two interesting problems:
1. When $ S=0 $ ?
2. Ascertain sum $ S\ if\ \left(\forall\right)k\in \overline {1,n},\ x_{k+1}=x_k+r $ (arithmetical progression).