Numbers $1, 2, 3, \ldots, 12$ are placed on all edges of a cube, one per edge. Then for each of the $8$ vertices of the cube, the sum of $3$ numbers converging on this vertex is computed. Is it possible all $8$ these sums are equal to each other?  

I started by adding $1$ to $12$ up and then multiplying by two since each number is counted twice on the cube. However, I am not sure how to prove that there exists a configuration since my process involves the assumption that it works.