any solution?

We will denote 'vertex triangle' for two triangle that precisely share only one side with other triangles, 'edge triangle' for $4(n-1)$ triangles that share two sides with other triangles, 'interior triangle' for remaining $2(n-1)^2$ triangles that share three sides with other triangles. The sum on edge can be always written in forms of $\sum a_i \times i$ where $a_i$ is $+1$ or $-1$ on vertex triangles, $2, 0$ or $-2$ on edge triangles and $3, 1, -1$ or $-3$ on interior triangles. Therefore it is always less than $3 \times (2n^2 + ... + n^2 +2n) + 2\times(n^2 + 2n-1 +...+n^2+2)+1\times (n^2+1)$ $ -1\times(n^2 ) -2\times (n^2 - 1 +...n^2 - 2n +2) - 3 \times (n^2-2n+1+ ...+1)$. Which is equal to $3n^4 - 4n^2 + 4n -2$. Corresponding example is easy to construct. Color the rhombus by chessboard coloring and place number greater than $n^2$ on black triangle and the others in white triangle. Place $n^2+1, n^2$ on vertex triangle, $n^2+2, ...n^2+2n-1$ and $n^2-1,...n^2-2n+2)$ on edge triangle and place the remaining in interior triangle. Therefore the answer is $3n^4 - 4n^2 + 4n -2$.