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 $24$?