The eleventh question: A certain meeting has 30 participants, each of whom has at most 5 acquaintances among the others, and in any group of 5 people, there exist two who are not acquaintances. Find the maximum positive integer $\mathrm{k}$, such that in the 30 people satisfying the above conditions, there always exist $\mathrm{k}$ people, none of whom are acquaintances with each other.
Question 11:
Solution Method One: Let the 30 people be $A_{1}, A_{2}, \ldots, A_{30}$. According to the conditions, $A_{1}$ is not familiar with at least 24 out of $A_{2}, A_{3}, \ldots, A_{30}$. Without loss of generality, assume $A_{1}$ is not familiar with $A_{2}, A_{3}, \ldots, A_{25}$.

According to the conditions, $A_{2}$ is not familiar with at least 18 out of $A_{3}, A_{4}, \ldots, A_{25}$. Without loss of generality, assume $A_{2}$ is not familiar with $A_{3}, A_{4}, \ldots, A_{20}$.

According to the conditions, $A_{3}$ is not familiar with at least 12 out of $A_{4}, A_{5}, \ldots, A_{20}$. Without loss of generality, assume $A_{3}$ is not familiar with $A_{4}, A_{5}, \ldots, A_{15}$.

According to the conditions, $A_{4}$ is not familiar with at least 6 out of $A_{5}, A_{6}, \ldots, A_{15}$. Without loss of generality, assume $A_{4}$ is not familiar with $A_{5}, A_{6}, \ldots, A_{10}$.

Finally, according to the conditions, among $A_{5}, A_{6}, \ldots, A_{10}$, at least two people are not familiar with each other. Without loss of generality, assume $A_{5}$ and $A_{6}$ are not familiar with each other. At this point, $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ are all not familiar with each other. Thus, we know that $k_{\text{max}} \geq 6$.
Finally, we need to construct a scenario where among any 7 people, there must be two who are familiar. We use 30 points $A_{1}, A_{2}, \ldots, A_{30}$ on a plane to represent the 30 people. If two people are familiar, we draw a line between the corresponding two points. The graph below meets the requirements:
In summary, the maximum value of $k$ that satisfies the conditions is 6.
Solution Method Two: We use 30 points $A_{1}, A_{2}, \ldots, A_{30}$ on a plane to represent the 30 people. If two people are familiar, we draw an edge between the corresponding two points, thus forming a graph $G$. According to the conditions, graph $G$ does not contain a complete subgraph of order 5, $K_{5}$, and $d(A_{i}) \leq 5, i=1, 2, \ldots, 30$. We need to find the largest integer $k$ such that the graph $G$ must contain a $k$-order independent set.

Let the maximum independent set of graph $G$ be $\{A_{1}, A_{2}, \ldots, A_{k}\}$, then the remaining $30-k$ points are each connected to at least one of these $k$ points. Since $d(A_{i}) \leq 5 (i=1, 2, \ldots, k)$, we have $5k \geq 30-k$, solving for $k \geq 5$.

If $k=5$, then $d(A_{i})=5 (i=1, 2, \ldots, 5)$, and each of $A_{1}, A_{2}, \ldots, A_{5}$ is connected to five points among $A_{6}, A_{7}, \ldots, A_{30}$. Without loss of generality, assume $A_{1}$ is connected to $A_{6}, A_{7}, \ldots, A_{10}$. Since graph $G$ does not contain a complete subgraph of order 5, there exist two points among $A_{6}, A_{7}, \ldots, A_{10}$ that are not connected. Without loss of generality, assume $A_{6}$ and $A_{7}$ are not connected, then $\{A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}\}$ is a 6-order independent set of graph $G$. This contradicts the maximality of $k=5$!
Thus, we know that $k_{\text{max}} \geq 6$.
Finally, we need to construct a graph $G$ that meets the conditions of the problem and does not contain a 7-order independent set. The graph below meets the requirements:
In summary, the maximum value of $k$ that satisfies the conditions is 6.