For $n\ge 6$, can we partition the set $\{1 , 4 , 9 , ...,n^2\}$ into two subsets whose sums are equal or differ by one? 
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?

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 $n\ge 6$.
