Dividing the squares $1^2,2^2,\ldots,54^2$ into three equal groups with the same total sum Is it possible to divide the squares $1^2,2^2,\ldots,54^2$ into three groups, each of which contains $18$ squares, such that the sum of squares within each group is the same for all three groups?

We can partition $9$ consecutive squares $n^2, (n+1)^2, \ldots,(n+8)^2$ in an almost equal fashion:
$$\begin{align}
(n+0)^2+(n+4)^2+(n+8)^2&=3n^2+24n+80\\
(n+1)^2+(n+5)^2+(n+6)^2&=3n^2+24n+62\\
(n+2)^2+(n+3)^2+(n+7)^2&=3n^2+24n+62\end{align} $$
Each group has three members and only one group has a sum too large by $18$.
By rotating the role of the too large partition, we can partition any $27$ consecutive squares into three groups of 9 members each and of equal sum.
As we can do this with $1^2,\ldots 27^2$ as well as $28^2,\ldots, 54^2$, the answer to the problem statement is: yes.
