I bet $10$ dollars that the answer will be $8$! case 1: $n$ is even, say $n=2k$, then we have that every language kills at most $\binom{k}{3}$ trios, so we must have: $14\binom{k}{3}$ is at least $\binom{2k}{3}$ and this only happens for $k>3$, so the smallest possible value is $8$, Igor told me this is the correct answer so i guess there is a nice configuration with it! case 2: $n$ is odd we get no better bound!

It is easy to construct the required sets using a 7 point projective plane + an extra point. Languages are spoken by lines + the extra point or by complements of lines.

remo wrote: It is easy to construct the required sets using a 7 point projective plane + an extra point. Languages are spoken by lines + the extra point or by complements of lines. Can you please explain? This sounds like a very cool method, but being a novice, I have no idea what you are talking about. Anyways, I did trial and error and found: 1234, 5678, 1256, 3478, 1278,3456, 1357,2468,1368,2457,1458,2367,1467, 2358. (1-8 denote the 8 people, and each of the 14 "clusters" gives the people who speak this language.) So indeed, the answer is 8.

Draw a triangle and mark it's vertices with 1,2,3. Draw its inscribed circle and mark the points where it touches the sides of triangle with 5,6,7 (so that 5 is on the side 2,3 and 6 on the side 1,3). Also mark the incenter with 4 and add one more point to that plane and mark it 8. Then read again the explanation and you'll see your example

I must be misunderstanding something. Isn't the only possible value of $n$ 8? Draw table with the 14 languages on the horizontal axis and the $n$ participants on the vertical axis. Put a marker in a cell if the participant speaks the corresponding language. The number of makrs is at least $n$ choose $3$, and at most $0.5*14n=7n$. This inequality is equivalent to $n \le 8$. What did I miss???

Wherefore have you infered the number of markers must be at least $\binom{n}{3}$ ? To better see this is not true, imagine there is no limitation on the number of persons speaking a same language, and have each person speaking all $14$ languages; then obviously any $3$ persons speak a common language, since they all three speak all languages. Then you have precisely $14n$ markers, and your inequality with $\binom{n}{3}$ is clearly misplaced.

Try double counting

Thusly I have herefore been proven wrong whence mavropnevma asked wherefore have I inferred that lower bound for the number of markers. I reach the obvious conclusion that I am thusforth incorrect insofar as the lowerbound so henceforth ignore my hitherto comment. Good day.