Partition of number's squares

Question

The problem is to divide $\{k^2\}_{k=1}^{1000}$ into two groups of 500 numbers each, such that they have equal sum.

I know that $$ \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}, $$ but it isn't enough for me to solve this. Please help me, if you can.

Answer 1

\begin{align} n^2+(n+3)^2&=2n^2+6n+9\;,\\ (n+1)^2+(n+2)^2&=2n^2+6n+5\;. \end{align}

So group the numbers in groups of $8$ and assign them $abbabaab$ in each group.