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This is not a number theoric solution!
Author of this problem should post a number theoric solution since this is categorized as number theory.

We can find all solutions using the following algorithm:

Note that $3465$ is half of the sum of all of the squares given by $\dfrac{27 \cdot 28 \cdot 55}{6} = 6930$ and hence sum of both $A$ and $B$ must be $3465$ .

1. Start with $27^2$ in $A$ .
2. If sum of $A$ is $3465$ we have a solution with $B$ containing the squares not in $A$ .
3. If $1^2$ is in $A$ remove it. If $A$ is empty, end the algorithm. Otherwise replace the smallest square in $A$ with the largest square smaller than it and go to step 2.
4. If sum of $A$ is greater than $3465$ , replace the smallest square in $A$ with the largest square smaller than it and go to step 2.
5. Add the largest square smaller than the smallest square in $A$ to $A$ and go to step 2.

The algorithm works because it starts with $27^2$ in $A$ and in each cycle it checks if the sum of the elements of $A$ is half of the sum of all of the squares and if it is we have a solution.

Then we continue to find the other solutions.
If we have $1^2$ in $A$ we cannot continue with it there so we remove it, end the algorithm if $A$ is empty and otherwise make the smallest square in $A$ smaller and go back to check if we have a solution.
Now if the sum is too large, since we know that we don't have $1^2$ in $A$ , we can just make the smallest square in $A$ smaller and go back to check if we have a solution.
Now we know that the sum is either less than or equal to the half of the sum of the squares, and since we know that we don't have $1^2$ in $A$ , we just add the largest square smaller than the smallest square in $A$ to $A$ and go check if we have a solution.

After only few cycles of this algorithm we find that,

$27^2 + \ldots + 23^2 + 17^2 + 5^2 + 4^2 = 22^2 + \ldots + 18^2 + 16^2 + \ldots 6^2 + 3^2 + 2^2 + 1^2$ .

Hence the answer is $\boxed{\text{Yes, it is possible}}$ .

I think it is little easy way to find A, B. By using " $(4k)^2 - (4k+1)^2 - (4k+2)^2 + (4k+3)^2 = 4$ "

By this way,

$24^2 - 25^2 - 26^2 + 27^2$

= $20^2 - 21^2 - 22^2 + 23^2$

= $16^2 - 17^2 - 18^2 + 19^2$

= $12^2 - 13^2 - 14^2 + 15^2$

= $8^2 - 9^2 - 10^2 + 11^2$

= 4

And, considering about " $(odd)^2 ≡ 1 (mod 8)$ "

$1, ~ 4, ~ 9, ~ 16, ~25, ~36, ~ 49 : 1, ~ 4, ~ 1, ~ 0, ~1,~ 4, ~ 1 (mod 8) ⇒ 1+9+25+49-4-16-36 > 20$ so unable

$1 ~4~ 9~ 16~ 25~ 36 ~49 ~64~ 81~ 100 ~121 : 1~ 4~ 1~ 0 ~1 ~4 ~1~ 0~ 1~ 4~ 1 (mod8)$ => $121-81-49-25-9-1 = -44 ( idea : 1-1-1-1-1 ≡ 4 (mod8) )$

so $-44 + 100 - 64 + 36 - 4 - 16 = 8$ and

$24^2 - 25^2 - 26^2 + 27^2$ - $20^2 - 21^2 - 22^2 + 23^2$ - $16^2 - 17^2 - 18^2 + 19^2$ - $12^2 - 13^2 - 14^2 + 15^2$ $= -8$

so we can find!!

$11^2 - 9^2 - 7^2 - 5^2 - 3^2 - 1^2 + 10^2 - 8^2 + 6^2 - 4^2 - 2^2 + 24^2 - 25^2 - 26^2 + 27^2$

$- ( 20^2 - 21^2 - 22^2 + 23^2 ) - ( 16^2 - 17^2 - 18^2 + 19^2 ) - ( 12^2 - 13^2 - 14^2 + 15^2 ) = 0$

Two more solutions:

First one is put 15 and all even numbers other than 6 into A, and 6 with all odd numbers other than 15 into B. That is:

15^2 + 2^2 + 4^2 + 8^2 + 10^2 + ... + 26^2 = 6^2 + 1^2 +3^2 + 5^2 + 7^2 + 9^2 + 11^2 + 13 ^2 + 17^2 + ... + 27^2

Second one is put 17 and all even numbers other than 10 into A, and 10 with all odd numbers other than 17 into B. That is:

17^2 + 2^2 + 4^2 + 6^2 + 8^2 + 12^2 + ... + 26^2 = 10^2 + 1^2 +3^2 + 5^2 + 7^2 + 9^2 + 11^2 + 13 ^2 + 15^2 + 19^2 + ... + 27^2

I realized that in order for A=B, both A and B need to be half of the sum of all numbers. I calculated (using an excel spreadsheet) that the sum is 6930 => A=B=6930/2=3465.

I then used the largest numbers , their sum not exceeding 3465, to find 27^2 + 26^2 + 25^2 + 24^2 + 23^2 = 529 + 576 + 625 + 676 + 729 = 3135 Then using the smaller numbers, I was able to find the missing 330; 2^2 + 6^2 + 11^2 + 13^2 = 4 + 36 + 121 + 169 = 330

From here it follows that the remaining numbers will also add up to 3465. Hence the answer is "Yes, it is possible".

A) 9 + 16 + 25 + 49 + 64 + 81 + 100 + 121 + 144 + 169 + 196 + 225 + 256 + 324 + 361 + 400 + 441 + 484
B) 1 + 4 + 36 + 289 + 529 + 576 + 625 + 676 + 729

4^2 + 8 2 + 16^2 + 23^2 + 24^2 + 25^2 + 26^2 + 27^2 = 3465 = (1/2) (6930) = sum from k = 1 to k = 27 of k^2. Ed Gray

dynamic programming for the win

I got it as squares of numbers from 5 to 20 and 25 add to 3465.

Omg..so many solutions exist...how many?

Denote A(n)=n (n+1) (2n+1)/6. Then A(22)-A(10)+A(5)=3465. But A(22)-A(10)+A(5)=1^2+...+5^2+11^2+...+22^2 and thus $$1^2+...+5^2+11^2+...+22^2=6^2+...+10^2+23^2+...+27^2$$.