There are $n$ participants at a conference, each knowing at most $k$ languages. For every group of three participants, there are at least two who speak the same language. Find the smallest value of $n$ such that for any distribution of languages satisfying these properties, one can find a language spoken by at least three delegates.
First, note that $n \geq 2k + 3$. Indeed, if $n = 2k + 2$, we can divide the participants into two groups of $k + 1$ people each. We take $2\binom{k+1}{2}$ different languages and distribute them among the pairs of people within the same group, so that two people from different groups cannot communicate. Whenever we take three people, there are two who belong to the same group and thus have a language in common.

Now, let's show that $n = 2k + 3$ works. Consider the graph $\tilde{A}$ with $2k + 3$ vertices, where the vertices are the participants of the conference, and the edges connect two participants who do not have a language in common. According to the problem statement, this graph is triangle-free. To ensure that three people speak the same language, we need to make "many" people speak the same language as a certain individual $A$, because then two of them will have a language in common due to the limited number of languages spoken by $A$. Therefore, we need to find a vertex of low degree in this graph. We know that there exists a vertex $A$ with degree $\left\lfloor\frac{2k + 3}{2}\right\rfloor \geq k + 1$. Thus, there are $k + 1$ participants $A_1, \ldots, A_{k+1}$ who can communicate with $A$. Since $A$ speaks at most $k$ languages, there must be two participants $A_i$ and $A_j$ who speak the same language among these $k$ languages. Therefore, $A, A_i, A_j$ have a language in common.

Remark. The arrangement for $n = 2k + 2$ is exactly the Turán graph $T(2k + 2, 2)$.