It is clear that if the system of linear equations
$$ \left\{ \begin{array}{c} x_1-x_3=c_1 \\ x_2-x_1=c_2 \\ x_3-x_2=c_3 \end{array} \right. $$
is solvable, then we have $c_1+c_2+c_3=0$.
How could we prove for the backward direction? That is, how could we prove that if $c_1+c_2+c_3=0$, then the above system of linear equations must be solvable without any corner cases?