For $n\ge 6$, can we partition the set $\{1 , 4 , 9 , ...,n^2\}$ into two subsets such that the sums of the elements in the two subsets are equal or differ by one?
For example : for $n = 10$, we can form the subsets $S_1 = \{100 , 64 , 25 , 4\}$ and $S2 = \{1 , 9 , 16, 36, 49, 81\}$. $S_1$ adds up to $193$ and $S_2$ adds up to $192$.
Can we also identify the elements that we can assign to individual subsets that satisfies this property?
On
Hagen von Eitzen's analysis can be extended to find subsets whose sums differ by exactly $1$, simply by noticing that $2^2 - 1^2 = 3.$
Thus, for example $1^2 + (3^2 + 6^2)$ must differ from
$2^2 + (4^2 + 5^2)$ by exactly $1$.
Having constructed this baseline example, in a manner (again) very similar to
Hagen von Eitzen's analysis, you can (for example) construct
$\{1,3,6,7,10,12,13\}$ and $\{2,4,5,8,9,11,14\}.$
The difference of sums is of the from $S:=\sum_{k=1}^n s_kk^2$ where each $s_k=\pm1$. Our task is to find $s_k$ such that the sum is $0$ or $1$.
Observe that $$\tag1a^2-(a+1)^2-(a+2)^2+(a+3)^2=4.$$ Therefore, we can pick four consecutive signs such that they contribute either $+4$ or $-4$. Hence with $8$ consecutive signs, we can achieve zero contribution. Here are the smallest non-negative sums we can achieve for some small $n$ with the eight distinct residues $\bmod 8$: $$ \begin{align}S_0&=0&=0\\ S_1&=1^2&=1\\ S_6 &= 1^2-2^2{+3^2-4^2-5^2+6^2}&=1\\ S_7 &= 1^2+2^2-3^2{+4^2-5^2-6^2+7^2}&=0\\ S_{10}&=-1^2+2^2-3^2-4^2{+5^2-6^2-7^2+8^2}-9^2+10^2&=1\\ S_{11}&=-1^2+2^2-3^2-4^2-5^2+6^2+7^2+8^2-9^2+10^2-11^2&=0\\ S_{12}&=-1^2-2^2-3^2-4^2-5^2-6^2-7^2-8^2+9^2+10^2-11^2+12^2&=0\\ S_{13}&=-1^2-2^2-3^2-4^2-5^2-6^2-7^2+8^2+9^2-10^2+11^2+12^2-13^2&=1 \end{align}$$ We conclude that we can reach sum $=0$ or $=1$ at least when $n$ si one of $0,1,6,7,10,11,12,13$ plus a multiple of $8$. In particular, this covers all $n\ge 6$.