I'm solving a problem regarding Newtonian forces. Since $F_{ij}=-F_{ji}$ then $\sum_j m_j\ddot{x}_j=0$. For a demonstration I was thinking to arrange all the $F_{ij}$ as a Matrix $F\in\mathbb{R}^{N\times N}$, including $F_{jj}=0$. If I could express the sum of all elements in such a matrix, as the sum of the diagonals $$F_{11}+(F_{12}+F_{21})+(F_{13}+F_{22}+F_{31})+\cdots+(F_{N\ N-1}+F_{N-1\ N})+F_{NN}$$ then the result follows immediately as each ($\cdots$) is zero.
Is there a formula to express $\sum_j\sum_i F_{ij}$ as exactly the sum above? I mean expressing as double sum or similar. Notice that each ($\cdots$) it's exactly the sum of all terms $F_{ij}$ such that $i+j=k$ for $k=2, \cdots, 2N$.
We sum up the elements of the matrix $(F_{i,j})_{1\leq i,j\leq N}$ by summing up the elements of the lower triangle matrix (marked $\color{blue}{\text{blue}}$), the main diagonal and the upper triangle matrix (marked $\color{red}{\text{red}}$).
We obtain \begin{align*} &\color{blue}{F_{1,1}+\left(F_{1,2}+F_{2,1}\right)+\left(F_{1,3}+F_{2,2}+F_{3,1}\right)}\\ &\qquad\color{blue}{+\cdots}\\ &\qquad\color{blue}{+\left(F_{1,N-1}+F_{2,N-2}+\cdots+F_{N-1,1}\right)}\\ &\qquad+F_{1,N}+F_{2,N-1}+\cdots+F_{N-1,2}+F_{N,1}\\ &\qquad\color{red}{+\left(F_{2,N}+F_{3,N-1}+\cdots+F_{N,2}\right)}\\ &\qquad\color{red}{+\cdots}\\ &\qquad\color{red}{+\left(F_{N-2,N}+F_{N-1,N-1}+F_{N,N-2}\right)+\left(F_{N-1,N}+F_{N,N-1}\right)+F_{N,N}}\\ &\,\,=\color{blue}{\sum_{k=1}^{N-1}\sum_{l=1}^{k}F_{l,k+1-l}}+\sum_{l=1}^{N}F_{l,N+1-l}+\color{red}{\sum_{k=2}^N\sum_{l=k}^NF_{l,N+k-l}} \end{align*}