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This is a recent problem in American Mathematical Monthly. The deadline for this question just passed:

$\textbf{Problem:}$ For which positive integers $n$ can $\{1,2,3,...,n\}$ be partitioned into two sets $A,B$ of the same size so that:

\begin{align}\sum_{k\in A}k&=\sum_{k \in B} k ,&\sum_{k\in A}k^{2}&=\sum_{k \in B} k^{2} , &\sum_{k\in A}k^{3}&=\sum_{k \in B} k^{3} \end{align}

It is clear that the set $\{1,2,...,16\}$ can be partitioned into $A=\{1,4,6,7,10,11,13,16\}$ and $B=\{2,3,5,8,9,12,14,15\}$, and the sum of the elements, its squares and cubes, are equal in $A$ and $B$. So for any $n$ divisible by $16$, an extension of this argument will work. It is not too difficult to show that $n$ has to be a multiple of $8$.

Can we have any $n$ which is an odd multiple of $8$, for which the problem statement is true; say $24$?

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    $\begingroup$ Works for all $n=8k$ with $k \geq 2$. See mat.uniroma2.it/~tauraso/AMM/AMM12085.pdf . $\endgroup$
    – Sil
    Jun 4, 2019 at 22:32
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    $\begingroup$ See also arxiv.org/pdf/1304.6756.pdf (Theorem 5.2): $A = \{ 1,3,7,8,9,11, 14, 16,17,18,22,24\}$ and $B$ the complement. $\endgroup$
    – Catalin Zara
    Jun 4, 2019 at 22:37
  • $\begingroup$ Wow, this is pretty interesting. Great post! I'm stumped but apparently @Sil knows what's up. $\endgroup$
    – C. Melton
    Jun 4, 2019 at 23:35
  • $\begingroup$ That is actually solution by user @RobertZ here on MSE, if I am not mistaken. $\endgroup$
    – Sil
    Jun 5, 2019 at 6:56
  • $\begingroup$ It's clear that once you show a partition for $n=16$ and for $n=24$, then for any multiple $8k, k\geq 2$ we get a required partition; this is outlined in the pdf. $\endgroup$
    – Aritro Pathak
    Jun 5, 2019 at 14:53

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