The 29th question: In a class, there are $\mathrm{n}$ students, and any two are either friends or not friends, with the friendship being mutual. In this class, any $\mathrm{n}-1$ students can be arranged in a circle such that any two adjacent students in the circle are friends, but all $\mathrm{n}$ students cannot be arranged in such a circle. Find the minimum value of $\mathrm{n}$.

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The 29th question: In a class, there are $\mathrm{n}$ students, and any two are either friends or not friends, with the friendship being mutual. In this class, any $\mathrm{n}-1$ students can be arranged in a circle such that any two adjacent students in the circle are friends, but all $\mathrm{n}$ students cannot be arranged in such a circle. Find the minimum value of $\mathrm{n}$.
The 29th question,
Answer: Represent students with points, and if two students are friends, connect the corresponding two points with an edge, to get the graph $G(V, E)$. The condition is: for any $v \in V(G), G-v$ is an $H$-graph, while $G$ itself is not an $H$-graph.

For any $v \in V(G)$, since $G-v$ is an $H$-graph and $G$ is not an $H$-graph, $v$ cannot be connected to two adjacent vertices on an $H$-cycle in $G-v$, hence $\mathrm{d}(\mathrm{v}) \leq\left[\frac{\mathrm{n}-1}{2}\right]$; on the other hand, if $\mathrm{d}(\mathrm{v}) \leq 2$, then removing a vertex $w$ adjacent to $\mathrm{v}$ from $\mathrm{G}$, $G-w$ does not contain an $H$-cycle. Therefore, for any $v \in V(G), 3 \leq d(v) \leq\left[\frac{n-1}{2}\right]$. Hence $\mathrm{n} \geq 7$.
If $n=7$, at this point the degree of each vertex can only be 3, thus $2|E|=\sum_{v \in V} d(v)=21$, which is a contradiction.
If $n=8$, at this point the degree of each vertex is still 3. Suppose $v_{1} v_{2} \ldots v_{7} v_{1}$ is an $H$-cycle in $G-v$, and assume without loss of generality that $v$ is connected to $v_{6}, v_{1}, v_{3}$, as shown in Figure (1). If $v_{5}, v_{7}$ are connected, then $v_{1} v_{2} v_{3} v_{4} v_{5} v_{7} v_{6} v$ is an $H$-cycle in $G$; if $v_{5}, v_{2}$ are connected, then $v v_{6} v_{7} v_{1} v_{2} v_{5} v_{4} v_{3} v$ is an $H$-cycle in $G$. This contradicts the condition.

If $\mathrm{n}=9$, then the degree of each vertex is 3 or 4, and they cannot all be 3. Thus, we can assume there exists a vertex $\mathrm{v}$ with degree 4, as shown in Figure (2).
Since $G-v_{2}$ contains an $H$-cycle $v v_{3} v_{4} v_{5} v_{6} v_{7} v_{8} v_{1} v$, $v_{2}$ is not connected to $v_{4}, v_{8}$; if $v_{2}$ is connected to $v_{6}$, then $v_{3} v_{4} v_{5} v_{6} v_{2} v_{1} v_{8} v_{7} v$ is an $H$-cycle in $G$, which contradicts the condition. Hence $v_{2}$ is connected to at least one of $v_{5}, v_{7}$. Similarly, each $v_{2i}$ is connected to at least one of $v_{2i+3}, v_{2i-3}$ (indices taken modulo 8). Since vertices with even indices are only connected to vertices with odd indices, removing a vertex with an odd index means the remaining graph cannot have an $H$-cycle, because the remaining graph does not have an equal number of vertices with odd and even indices.
Finally, provide a graph for $n=10$ that satisfies the condition. As shown in Figure (3), the Petersen graph itself is not an $H$-graph, but removing any vertex results in an $H$-graph.

In summary, the smallest value of $\mathrm{n}$ that satisfies the condition is 10.