Seems closely related to All-Russian 2005 9.8 & 11.8. @below Ok, actually you're right.

@above I don't think so. It is correct that a) requires a greedy algorithm (though much simpler than the ARO problem) and for b) I doubt that anything of this sort would work.

This nice problem was proposed by Miroslav Marinov. Part a) is just for warm up. Obviously, $k=1$ is not enough and it's possible for $k=2$. The worst case scenario is when all the students have the same favorite numbers, say $1$ and $2$. In this case we arrange the numbers as $1-2-1-\dots 2$. Part b). $k=4$. I follow in general the author's solution. We omit the counterexample that it is not possible for $k=3$ and let's prove that it's always possible when $k=4$. Consider a bipartite graph $G(S, N)$ where $S=\{1,2,\dots,14\}$ are the students and $N=\{n_1,n_2,\dots,n_k\}$ are all the numbers the students like. A vertex (student) $s\in S$ is connected with $n_i\in N$ if $n_i$ is a favorite number of $s$. The given condition says that every $s$ is connected with at least $4$ vertices in $N$, i.e. $d(s)\ge 4, \forall s\in S$. Let us partition the vertices $S$ into two groups $S_0$ (boys) and $S_1$ (girls). The problem actually asks to prove the following claim. $\textbf{Claim.}$ The vertices in $N$ can be partitioned into two disjoint sets $N_0$ and $N_1$, such that each vertex in $S_0$ is connected with a vertex in $N_0$ and each vertex in $S_1$ is connected with a vertex in $N_1$. $\textit{Proof.}$ Each partition of $N$ into two sets can be interpret as assigning to each $n_i\in N$ either a value $0$ if $n_i\in N_0$ or $1$ in case $n_i\in N_1$. Clearly the family $\mathcal{A}$ of all assignments (partitions) consists of $2^k$ elements. For each $s\in S$ let us denote by $B_s\subset \mathcal{A}$ the set of "bad" assignments for the student $s$. That is, in case $s\in S_0$, $B_s$ consists of all assignments for which to all neighbors of $s$ is assigned $1$, and in case $s\in S_1,$ $B_s$ are those assignments for which $s$ is connected only with $0$'s. It means $$|B_s|= \frac{2^k}{2^d}$$where $d=d(s)$. Since $d(s)\ge 4$ we get $|B_s|\le 2^{k-4}$. Let $B:=\bigcup_{s\in S}B_s$ be all bad assignments. It yields $$|B|=\left|\bigcup_{s\in S}B_s\right|\le \sum_{s\in S}|B_s|\le 14\cdot 2^{k-4}<2^k.$$Therefore $|B|<|\mathcal{A}|$ which means there exists an assignment $A\in \mathcal{A}\setminus B$. Obviously, $A$ comply with all the requirements of the Claim. $\textbf{Comment}.$ Actually, it's a disguised probabilistic approach (or vise versa )Note that in the above proof it's not essential the students are $7$ boys and $7$ girls. The size of $S_0$ and $S_1$ can be whatever we want, providing $|S_0\cup S_1|=14$. Moreover, the same holds even if the number of students is $16$. But, in this case a further argument is needed. Namely, for any $i,j\in S$ that are of the same sex, that is either $i,j\in S_0$ or $i,j\in S_1$, it holds $B_i\cap B_j\neq \emptyset$.