5. A certain meeting has 30 participants, each of whom knows at most five others; among any five people, at least two are not acquaintances. Find the largest positive integer $k$, such that in a group of 30 people satisfying the above conditions, there always exists a subgroup of $k$ people, none of whom are acquaintances.
（Leng Gangsong, Contributed）
5. Represent 30 people with 30 points. If two people are acquaintances, connect a line between their corresponding points, thus forming a simple graph $G$ with these 30 points as the vertex set, satisfying the following conditions:
(i) The degree of each vertex in graph $G$ does not exceed 5;
(ii) Among any five points in the vertex set of graph $G$, there are two points not connected by an edge.

Let $V$ denote the vertex set of graph $G$. If $A \subseteq V$, and any two points in set $A$ are not connected by an edge, then $A$ is called an "independent set" of graph $G$, and the independent set with the maximum number of elements is called a "maximum independent set" of graph $G$.
(1) First, prove: In a graph $G$ that satisfies the conditions, the number of elements in the maximum independent set is no less than 6.

In fact, let $X$ be a maximum independent set of graph $G$. By the maximality of $|X|$, we know that any point in the set $V \backslash X$ has a neighbor in the set $X$. Otherwise, if $a \in V \backslash X$ has no neighbor in the set $X$, then we can add $a$ to the set $X$ to form a larger independent set, which is a contradiction. Thus, there are at least $|V \backslash X| = 30 - |X|$ edges between the set $V \backslash X$ and $X$.

Also, note that the degree of each point in the set $X$ does not exceed 5, so we have
$30 - |X| \leq 5|X|$.
Therefore, $|X| \geq 5$.
If $|X| = 5$, then by the equality in (1), we know that the $30 - |X| = 25$ edges are distributed among the five vertices in the set $X$, meaning that the neighbor set of each point in the set $X$ is a set of five points in the set $V \backslash X$.

Since $|V \backslash X| = 25$, the neighbor sets of any two points in the set $X$ do not intersect.
Let $X = \{a, b, c, d, e\}$.
Now consider the neighbor set of $a$, denoted as
$$
Y_{a} = \{y_{1}, y_{2}, y_{3}, y_{4}, y_{5}\}.
$$

By (ii), we know that there are two points in the set $Y_{a}$ that are not connected by an edge, say $y_{1}, y_{2}$.

Since the neighbor sets of any two points in the set $X$ do not intersect, $y_{1}, y_{2}$ cannot be neighbors of $b, c, d, e$.

Thus, $\{y_{1}, y_{2}, b, c, d, e\}$ is an independent set of graph $G$ with more than 5 elements, which is a contradiction.
This proves that $|X| \geq 6$.
(2) Next, prove: There exists a graph $G$ that satisfies the conditions, and the number of elements in its maximum independent set does not exceed 6.
Partition the set $V$ into point sets $V_{1}, V_{2}, V_{3}$, such that
$|V_{i}| = 10 (i=1,2,3)$.
Let $V_{1} = \{A_{1}, A_{2}, \cdots, A_{5}, B_{1}, B_{2}, \cdots, B_{5}\}$, and connect the points in the set $V_{1}$ as shown in Figure 3, i.e.,
(i) Connect $A_{i}A_{i+1} (i=1,2, \cdots, 5)$;
(ii) Connect $B_{i}B_{i+1} (i=1,2, \cdots, 5)$;
(iii) Connect $A_{i}B_{i}, A_{i}B_{i+1}, A_{i}B_{i-1} (i=1,2, \cdots, 5)$,
where $A_{6} = A_{1}, B_{6} = B_{1}, B_{0} = B_{5}$.
The connection method for point sets $V_{2}, V_{3}$ is the same as that for point set $V_{1}$, and for any $1 \leq i < j \leq 3$, there are no edges between point sets $V_{i}$ and $V_{j}$. Then, the degree of any vertex in graph $G$ is 5, and among any five points in graph $G$, there are always two points not connected by an edge.
Now, take any maximum independent set $X$ of graph $G$.
We only need to prove: $|V_{1} \cap X| \leq 2$.
In fact, since $A_{i}, A_{i+1} (i=1,2, \cdots, 5)$ are adjacent, at most two of $A_{1}, A_{2}, \cdots, A_{5}$ belong to set $X$.

Similarly, at most two of $B_{1}, B_{2}, \cdots, B_{5}$ belong to set $X$.

If exactly two of $A_{1}, A_{2}, \cdots, A_{5}$ belong to set $X$, say $\{A_{1}, A_{3}\}$.

Note that the union of the neighbor sets of $A_{1}$ and $A_{3}$ is exactly $\{B_{1}, B_{2}, \cdots, B_{5}\}$. Therefore, $B_{1}, B_{2}, \cdots, B_{5}$ do not belong to set $X$.

Similarly, if exactly two of $B_{1}, B_{2}, \cdots, B_{5}$ belong to set $X$, then $A_{1}, A_{2}, \cdots, A_{5}$ do not belong to set $X$. This proves that $|V_{1} \cap X| \leq 2$.
Similarly, $|V_{2} \cap X| \leq 2, |V_{3} \cap X| \leq 2$.
Thus, $|X| = |V \cap X|$
$$
= |V_{1} \cap X| + |V_{2} \cap X| + |V_{3} \cap X| \leq 6.
$$

Therefore, graph $G$ meets the requirements.
Combining (1) and (2), we know that the value of $k$ is 6.