i had earlier written a wrong solution but i realized it when i was working on something else. this time i think i have made amends. the answer is $96$. to see that $n\geq 96$ we can partition the numbers till $95$ as follows: $\{1,2,3,4,5,7,9\}\cup\{48,60,66,\ldots,90,80,88\},\{10,11,12,\ldots,95\}\cup\{6,8\}\setminus\{60,66,\ldots,90,80,88\}$. to see that $96$ is sufficient, we basically focus our attention on the numbers $2,3,4,6,8,12,24,48,96$.one can argue (easily) taking three cases as to which of $2,4,8$ is the first element among these to co-occur with $3$(if they all are on the other partition from $3$ then of course the problem is solved).i'll write one part-the others are as easy.if $2,3$ co-occur then if $4$ also occurs in the same part then necessarily $6,8,12$ occur in the other part. then $48,96$ must occur along with $2$ and that is a desired triple. if $4$ does not occur along with $2,3$ then $4,6$ occur in the other partition so necessarily $24$ occurs along with $2,3$ which then implies that $48$ occurs in the other part. but this again forces $8$ into this same partition and hence we have the triple $3,8,24$ as desired, and so on. the only thing is that this solution is not satisfactory and possibly there might be a nicer way of seeing things.