Solution:

Call a run any set of four consecutive two-digit integers containing a multiple of $6$ and a different multiple of $7;$ the runs are exactly the candidates for $S$ allowed by the announcement. A student holding a number that lies in exactly one run could name $S$ immediately, so the unanimous "no" reveals that every element of $S$ lies in at least two runs.

A number belongs to two different runs only when nearby runs overlap, which happens when a multiple of $6$ and a multiple of $7$ are consecutive integers, both two-digit: the pairs $(35, 36),$ $(48, 49),$ $(77, 78),$ and $(90, 91).$ Checking each cluster, the runs all four of whose elements are ambiguous are exactly the ones with such a pair in the two middle positions: $$\{34, 35, 36, 37\}, \quad \{47, 48, 49, 50\}, \quad \{76, 77, 78, 79\}, \quad \{89, 90, 91, 92\}.$$ These four sets are pairwise disjoint, so after the four "no" replies each student's own number singles out one of them, consistent with everyone then deducing $S.$

The possible greatest elements are $37,$ $50,$ $79,$ and $92,$ with sum $258.$