(a) Since each student can choose at most $k$ numbers, there are at most $nk$ free connections. However, in total, there are $\frac{n(n-1)}2\ge\frac{n(2k+1)}2>nk$ connections among the students, so one of them will not be free. (b) Label the students $0,1,\ldots,2k$. $i$ chooses to speak for free with $j$ iff $j-i\in\{1,2,\ldots,k\}\pmod{2k+1}$. This arrangement works.

In graph-theoretical terminology a) No digraph on $ n \geq 2k+2$ vertices and maximal outdegree at most $ k$ can have $ K_n$ as its subjacent unoriented graph; b) Among all digraphs on $ n = 2k+1$ vertices and maximal outdegree at most $ k$ there exists one having $ K_n$ as its subjacent unoriented graph. Notice in the proof above that in case a) not just one, but at least $ \frac {n} {2}$ connections will not be free. The model presented for case b) can be prolonged for $ n=2k+2$ to ensure that no more than $ \frac {n} {2} = k+1$ connections are not free.