2 A meeting has $12 k$ people attending, each person has greeted exactly $3 k+6$ people. For any two people, the number of people they have greeted is the same. How many people attended this meeting?
2. Represent people with dots, and for two people who have greeted each other, connect the corresponding two dots to form a simple graph. Calculate the number of angles $S$ in this graph. On one hand, for $G$, each point leads to $3k+6$ edges, resulting in $\mathrm{C}_{3k+6}^{2}$ angles. With $12k$ points, there are $12k \mathrm{C}_{3k+6}^{2}$ angles in total. Since each angle has a vertex and different vertices correspond to different angles, $S=12k \mathrm{C}_{3k+6}^{2}$. On the other hand, starting from a "pair of people," since the number of people who have greeted any two people is the same, let this number be $t$, then each pair of people contributes $t$ angles. Since each angle corresponds to a pair of people, and different pairs of people correspond to different angles, $S=t \mathrm{C}_{12k}^{2}$. Therefore, $12k \mathrm{C}_{3k+6}^{2}=t \mathrm{C}_{12k}^{2}$, solving for $t$ gives $t=\frac{(3k+6)(3k+5)}{12k-1}$. Thus, $16t=\frac{(12k-1+25)(12k-1+21)}{12k-1}=(12k-1)+25+21+\frac{25 \times 21}{12k-1}$. Clearly, $(3, 12k-1)=1$, so $12k-1 \mid 25 \times 7$. Noting that $12k-1$ is congruent to 3 modulo 4, so $12k-1=7$, $5 \times 7$, $5^2 \times 7$, among which only $12k-1=5 \times 7$ has an integer solution $k=3$, $t=6$, so the number of people at the meeting can only be 36. Below, we prove that a meeting of 36 people is possible. That is, there exists a 36-vertex graph $G$ where each vertex has a degree of 15, and for every pair of vertices, there are 6 vertices connected to both. First, construct 6 complete graphs $K_{6}$. Each complete graph's vertices are numbered $1, 2, 3, 4, 5, 6$. Now connect the relevant points of these 6 complete graphs to form the graph $G$. The vertices in $G$ are denoted as $(i, j)$, representing the $j$-th vertex in the $i$-th complete graph. Note that each vertex is already connected to 5 vertices within its own complete graph. Now connect each vertex to the vertices with the same second coordinate, adding 5 more edges. Then connect vertices with the same coordinate difference, i.e., $i-j=i^{\prime}-j^{\prime}$, so $(i, j)$ and $\left(i^{\prime}, j^{\prime}\right)$ are connected. This way, each vertex has 15 edges, and each vertex is connected to 2 vertices in each of the other complete graphs, one with the same second coordinate and one with the same coordinate difference. For any two vertices, if a vertex is connected to both, it is said that these two vertices form an angle. Below, we prove that any two vertices in $G$ form 6 angles. In fact, consider any two vertices $(i, j)$ and $\left(i^{\prime}, j^{\prime}\right)$. If $i=i^{\prime}$, i.e., these two vertices are in the same complete graph, then they form 4 angles within that complete graph. Additionally, exactly two vertices: $\left(i^{\prime}-j^{\prime}+j, j\right)$ and $\left(i-j+j^{\prime}, j^{\prime}\right)$ are connected to both, so they form 6 angles in total. If $i \neq i^{\prime}$ and $i-j=i^{\prime}-j^{\prime}$, then $\left(i, j^{\prime}\right)$ and $\left(i^{\prime}, j\right)$ are connected to both. Additionally, for the 4 $i^{\prime \prime}$ outside of $i$ and $i^{\prime}$, the vertices $\left(i^{\prime \prime}, i^{\prime \prime}-i+j\right)$ are connected to both, and no other vertices are connected to both. So they form 6 angles. If $i \neq i^{\prime}$ and $i-j \neq i^{\prime}-j^{\prime}$, then $\left(i, j^{\prime}\right)$, $\left(i^{\prime}, j\right)$, $\left(i, i-i^{\prime}+j^{\prime}\right)$, and $\left(i^{\prime}, i^{\prime}-i+j\right)$ are connected to both. Additionally, the vertices $\left(i^{\prime}-j^{\prime}+j, j\right)$ and $\left(i-j+j^{\prime}, j^{\prime}\right)$ are connected to both, and no other vertices are connected to both. So they form 6 angles. In summary, the number of people at the meeting is 36.