For which values of positive integer k is it possible to divide the first 3k positive integers into three groups with the same sum? I'm on a GCSE-a level syllabus currently, and I can't seem to think of any algebraic equation that I could comprise to solve this (with the GCSE/early a level syllabus). The question in full is

For which values of positive integer k is it possible to divide the first 3k positive integers into three groups with the same sum?
(e.g. if k = 3, then the first 3k integers are 1,2,3,4,5,6,7,8,9. You can split these into 3 groups of 15, for example {{1,2,3,4,5},{7,8},{6,9}}. so it is possible for k=3)

Any help would be appreciated. Thanks

The sum of the first $3k$ numbers is $\frac 12(3k)(3k+1)$, so we want three groups that sum to $\frac 12k(3k+1)$.  Clearly $k=1$ doesn't work because the desired sum is $2$ and we have a $3$ which is too big.  $k=2$ does work as we have $7=1+6=2+5=3+4$.  Intuitively, as we have more numbers we have more freedom, so expect it to work.  You have shown a solution for $k=3$.  Now note any number greater than $1$ can be expressed as a sum of $2$s and $3$s, so we can separate $3k$ into a sum of runs of $6$ and $9$.  Our examples have the feature that each group is the same size, so we can add a constant to each entry and get a division of the numbers from $m$ to $m+5$ or $m+8$ into three groups.  Each run can be treated as one of our examples, so the problem can be solved for all $k \gt 1$
