# Cube problem with twice 2016

There are cubes with side lengths $1,2,3,...,N$ made from the same metal material, so that their weights are the the cube integers $$1^3, 2^3, 3^3, .... ~ ...., (N-1)^3, N^3.$$ The cubes can be divided into two groups so that the heavier group weights exactly $20162016$ more than the lighter group.

Is there exists an integer $N \geq 1$ for which this story is possible?

let $N=95$. The lighter group consists of $2^3$, $6^3$ through $12^3$, and $21^3$ through $34^3$, and has a total weight of $315792$. The heavier group contains all the other cubes, and has a weight of $20477808$. The difference is $20162016$, as desired.
Alternatively, notice that for any $n$,
$$(n+7)^3-(n+6)^3-(n+5)^3+(n+4)^3-(n+3)^3+(n+2)^3+(n+1)^3-n^3=48$$
so out of any eight consecutive cubes, they can be divided into two groups where the heavier group weighs $48$ more than the lighter one. $20162016$ is divisible by $48$, so we can just keep adding eight cubes in this pattern until we reach it.