8. There are $n$ mathematicians attending a conference, and each mathematician has exactly 3 friends (friendship is mutual). Now, if we arrange some of the mathematicians to sit around a round table to discuss problems, it turns out that if everyone is seated next to their friends, then the number of people at the table is no less than 7. Find the minimum possible value of $n$.
Analysis: Consider mathematicians as vertices, and if two mathematicians are friends, connect an edge between the corresponding vertices, thus obtaining a simple graph $G$. According to the problem, graph $G$ has $n$ vertices, each vertex has a degree of 3, and the length of each cycle $\geq 7$. Since each vertex in $G$ has a degree of 3, the sum of the degrees of all vertices is $3n$, which equals twice the number of edges in $G$, so $n$ is even.

Take any vertex $A$ in $G$, and let its three neighbors be $A_{1}, A_{2}, A_{3}$. At this point, $A_{i}$ has two other neighbors, denoted as $A_{i1}, A_{i2}$; and $A_{ij}$ has two other neighbors $A_{ij1}, A_{ij2}$, $1 \leq i \leq 3, 1 \leq j \leq 2$. It is easy to see that these $1+3+6+12=22$ vertices are all distinct (otherwise, there would be a cycle of length $\leq 6$). Therefore, $n \geq 22$.

If $n=22$, we observe the subgraph $G^{*}$ formed by the 12 points $A_{ijk}$ (where $1 \leq i \leq 3, 1 \leq j, k \leq 2$). Each vertex in $G^{*}$ has a degree of 2, so $G^{*}$ is the union of several cycles. But each cycle has a length $\geq 7$, so $G^{*}$ must be a cycle of length 12. Moreover, the points in $G^{*}$ can be divided into 6 pairs, each pair having a common neighbor in $G \backslash G^{*}$.

Now, relabel the points in $G^{*}$ in the order of the cycle as $B_{1} B_{2} \cdots B_{12}$. Suppose $B_{1}$ and $B_{a}$ have a common neighbor $C_{1}$ in $G \backslash G^{*}$, $B_{2}$ and $B_{b}$ have a common neighbor $C_{2}$, $B_{3}$ and $B_{c}$ have a common neighbor $C_{3}$, and $B_{4}$ and $B_{d}$ have a common neighbor $C_{4}$. Since each cycle in $G$ has a length $\geq 7$, it is easy to see that $6 \leq a \leq 8$, $7 \leq b \leq 9$, $8 \leq c \leq 10$, $9 \leq d \leq 11$, and $|b-c| \geq 2$, $|c-d| \geq 2$. From the five inequalities about $a, b, c$, we can deduce that $a=6$, $b=8$, $c=10$. At this point, $|c-d| \leq 1$, which contradicts $|c-d| \geq 2$.

Therefore, $n \geq 24$. An example where the equality holds is as follows: First, construct 3 cycles of length 8: $A_{1} A_{2} \cdots A_{8}$, $B_{1} B_{2} \cdots B_{8}$, $C_{1} C_{2} \cdots C_{8}$. Then connect the following edges: $A_{1} B_{1}$, $A_{2} B_{5}$, $A_{5} B_{2}$, $A_{6} B_{6}$, $A_{3} C_{3}$, $A_{4} C_{7}$, $A_{7} C_{4}$, $A_{8} C_{8}$, $B_{3} C_{1}$, $B_{4} C_{5}$, $B_{7} C_{2}$, $B_{8} C_{6}$. It is easy to verify that each vertex has a degree of 3, and the smallest cycle length is 7. Therefore, the minimum value sought is 24.