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 any group of 30 people satisfying the above conditions, there always exists a group of $k$ people, none of whom are acquaintances.
(Cold Gangsong, Contributed)
5, hence we have
$30-|X| \leqslant 5|X|$.
Therefore, $|X| \geqslant 5$.
If $|X|=5$, then by equation (1) the equality is achieved, which means the 25 edges are distributed among the five vertices in set $X$, i.e., the neighborhood of each vertex in set $X$ is a set of five points in set $V \backslash X$.

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

By (ii), we know that there are two points in set $Y_{a}$ that are not connected, let them be $y_{1}$ and $y_{2}$.

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

Thus, $\left\{y_{1}, y_{2}, b, c, d, e\right\}$ is an independent set of graph $G$ with more than 5 elements, which is a contradiction.
This proves that $|X| \geqslant 6$.
(2) Next, we prove: there exists a graph $G$ that meets the conditions, and the maximum independent set has no more than 6 elements.
Partition the set $V$ into point sets $V_{1}, V_{2}, V_{3}$, such that
$\left|V_{i}\right|=10(i=1,2,3)$.
Let $V_{1}=\left\{A_{1}, A_{2}, \cdots, A_{5}, B_{1}, B_{2}, \cdots, B_{5}\right\}$, connect the points in set $V_{1}$ as shown in Figure 3, i.e.,
(i) $A_{i} A_{i+1}(i=1,2, \cdots, 5)$ are connected;
(ii) $B_{i} B_{i+1}(i=1,2, \cdots, 5)$ are connected;
(iii) $A_{i} B_{i}, A_{i} B_{i+1}, A_{i} B_{i-1}(i=1,2, \cdots, 5)$ are connected, where $A_{6}=A_{1}, B_{6}=B_{1}, B_{0}=B_{5}$.

The connection method for point sets $V_{2}$ and $V_{3}$ is the same as for set $V_{1}$, and for any $1 \leqslant i<j \leqslant 3$, there are no edges between point sets $V_{i}$ and $V_{j}$. Then, the degree of any vertex in graph $G$ is 5, and in graph $G$, among any five points, there are always two points that are not connected.
Now, take any maximum independent set $X$ of graph $G$.
We only need to prove: $\left|V_{1} \cap X\right| \leqslant 2$.
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 $X$, let them be $\left\{A_{1}, A_{3}\right\}$.

Notice that, the union of the neighborhoods of $A_{1}$ and $A_{3}$ is exactly $\left\{B_{1}, B_{2}, \cdots, B_{5}\right\}$. Hence, $B_{1}, B_{2}, \cdots, B_{5}$ 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 $\left|V_{1} \cap X\right| \leqslant 2$.
Similarly, $\left|V_{2} \cap X\right| \leqslant 2,\left|V_{3} \cap X\right| \leqslant 2$.
Thus, $|X|=|V \cap X|$
$$
=\left|V_{1} \cap X\right|+\left|V_{2} \cap X\right|+\left|V_{3} \cap X\right| \leqslant 6.
$$

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