Split the set of vertices into 2 sets $A$ and $B$ such that $0 \leq |A| - |B| \leq 1$ (in particular, $A = \lceil \frac{n}{2} \rceil$ and $B = \lfloor \frac{n}{2} \rfloor$ . Then, for any edge $e = xy$ where $x \in A$ and $y \in B$, assign an integer less than or equal to $|A||B| = \lfloor \frac{n^2}{4} \rfloor$. Then, assign any of the remaining value to all the other edges. Now, for any triangle, at least one of the edges of the triangle joins two vertices, where either both of them are in $A$ or both are in $B$. The number assigned to this edge is at least $|A||B| + 1 = \lfloor \frac{n^2}{4} \rfloor + 1 > 3n$, so the sum of all the numbers is strictly greater than $3n$.