Given 30 students such that each student has at most 5 friends and for every 5 students there is a pair of students that are not friends, we need to determine the maximum $k$ such that for all such possible configurations, there exists $k$ students who are all not friends.

In graph theory terms, we are given a regular graph with 30 vertices and degree 5, with no $K_5$ subgraphs. We aim to find the maximum size $k$ of an independent set in such a graph.

We claim that $k = 6$. To show this, we need to construct a graph that satisfies the given conditions and has an independent set of size 6, and also prove that any such graph must have an independent set of at least size 6.

Consider a graph $G$ with 10 vertices: $v_1, v_2, v_3, v_4, v_5, w_1, w_2, w_3, w_4, w_5$. Construct two cycles $v_1v_2v_3v_4v_5$ and $w_1w_2w_3w_4w_5$, and for $i, j \in \{1, 2, 3, 4, 5\}$, join $v_i$ and $w_j$ if and only if $i - j \equiv 0, \pm 1 \pmod{5}$. This graph $G$ has no independent set of size greater than 2 and no $K_5$.

Now, consider a graph $G'$ that consists of three copies of $G$. The maximum size of an independent set in $G'$ is no more than three times the maximum size of an independent set in $G$, which is 6. Thus, $G'$ is a $K_5$-free regular graph with degree 5 and an independent set of size at most 6.

To show that any graph satisfying the conditions has an independent set of size 6, we use Turán's Theorem. The complement graph $\overline{G}$ has 30 vertices and at least 360 edges. If $\overline{G}$ does not have a $K_6$, then by Turán's Theorem, $G$ can have at most 360 edges, leading to a contradiction. Therefore, $\overline{G}$ must have an independent set of size 6, implying $G$ has an independent set of size 6.

Thus, the maximum $k$ such that there exists $k$ students who are all not friends is:
$$\boxed{6}$$