Chatting in Olympiads

I can prove explicitly that the answer is at least 3: by the pigeonhole principle , one of the students corresponds with at least six students on the same topic, say topic A. Then if any pair of those six students corresponds on topic A, there is a group of three--otherwise, all six of those students correspond with each other on the other two topics (say B and C). But now we've got the well-known result from Ramsey theory that $R(3,3) = 6$ : take one of the six students and again notice that he corresponds with at least three students on the same topic (by Pigeonhole again). Say it's B. If any pair of those three correspond on B, there is a group of three, and otherwise those three all correspond on C.

To show that the answer is not 4 would require us to exhibit a 3-coloring of $K_{17}$ with no monochromatic $K_4.$ In fact, there is a 2-coloring of $K_{17}$ with no monochromatic $K_4$ ! That is, $R(4) = 18.$ This is obviously much more powerful than we need, but I thought it was pretty so I'll give the proof: take $17$ points spaced equally around the unit circle and color the edges red if the points are $1,2,4,$ or $8$ apart, and blue if the points are $3,5,6,$ or $7$ apart. Now convince yourself that there is no monochromatic $K_4$ in that picture.