Well i personally believe that:
This procedure can be done for all n. >!Correct me if im wrong please.
Something we will need later:
T1: Two perfect squares differ by >!atleast 3. Proof: x >=1: (x+1)^2-x^2= >!2x+1 >= 3
Ok the base my proof is:
I will use induction on n. The >!induction base can be n=9 as in the >!question. In the induction step we >!know that the hypothesis is true for >!n=k-1 now we need to prove the >!hypothesis for n =k.
Ok now the proof begins:
Now we construct our 3 sets for >!n=k-1. Now we name the sets :A,B,C. >!Now if the numbers 2 and 1 are in A >!and we add k to A, and S(A) becomes >!a square according to T1 if we take 2 >!or 1 S(A) will be not a squre and if we >!add for exmaple 2 to another set and >!it doesnt become a square then its ok >!otherwise we dont pick 2 and we pick >!1. In the case which only one of 1, 2 >!are in A(say x is that number), if when >!adding k to A, the sum becomes a >!square we can take x and put it in the >!set where the other of 1,2 is if that set >!becomes a square we can just add >!the other number back to A. Now if >!both of 1,2 are in the same set other >!than A, if when adding k to A its sum >!becomes a square we can give 1 to A >!if the other set becomes a square too >!we can switch 1 and 2. Now the last >!case: if 1,2 are in separate sets other >!than A, we can do the above >!procedures for another set like B >!because this case ia covered. This >!solution (all of it) depends on T1.
I hope my answer is atleast partially true.