3. Given a positive integer $n \geq 3$. Several people attend a meeting, some of whom know each other. If some people can sit around a round table such that adjacent people all know each other, this is called a "good" seating arrangement. It is known that these people cannot be divided into $n$ groups such that no one in each group knows anyone else. Find the largest positive integer $N$, such that there must be at least one person who can participate in $N$ different "good" seating arrangements.
※A "good" seating arrangement can consist of 1 person or 2 people who know each other, and any seating arrangements that are equivalent by rotation are considered the same.
【Answer】 $\max N=[e \cdot n!]$. (3 points)
Describe the problem using graph theory language, treating people as vertices and connecting edges between those who know each other. The condition of the problem then becomes that the graph cannot be colored with $n$ colors such that adjacent vertices have different colors, and the problem is to study the number of simple cycles passing through a certain vertex.

Take a complete graph $K_{n+1}$, which meets the requirements of the problem. However, the number of valid seating arrangements for any vertex inside it is the permutation of all other $k(0 \leq k \leq n)$ vertices, i.e., $1+n+n(n-1)+n(n-1)(n-2)+\ldots+n!+n!=n!\sum_{i=0}^{n} \frac{1}{i!}=[e \cdot n!]$ kinds. (6 points)
【Explanation】Let this graph be $G$, and let $H$ be the subgraph of $G$ with the maximum number of vertices that can be colored with $n$ colors. Since $G$ cannot be colored with $n$ colors, there is at least one vertex $A$ outside, and it has an edge connecting to the graph $H$, denoted by its neighborhood as $N(A)$. (9 points)

We take a coloring method of $H$ with $n$ colors. For any color sequence: $c_{1}, c_{2}, \ldots, c_{k}$, we divide some vertices of $H$ into $k$ sets $V_{1}, V_{2}, \ldots, V_{k}$ according to the following steps:
(1) Place the $c_{1}$-colored vertices in $N(A)$ into $V_{1}$, which is obviously possible, otherwise, $A$ can be colored with $c_{1}$.
(2) Place the $c_{2}$-colored vertices adjacent to vertices in $V_{1}$ into $V_{2}$.
(3) Continue this process until placing the $c_{k}$-colored vertices adjacent to vertices in $V_{k-1}$ into $V_{k}$.

If there are still $c_{1}$-colored vertices adjacent to vertices in $V_{k}$, place them into $V_{1}$ and repeat the above operation until it can no longer be performed.
Since the number of vertices is finite, such an operation will always stop. (12 points) We now state and prove several facts:
(1) $V_{k}$ is non-empty.

Assume $V_{k}$ is empty, then the operation stops in the first cycle, which means there exists $V_{i}(i<k)$ such that no $c_{i+1}$-colored vertices are adjacent to them. We change the color of vertices in $V_{i}$ to $c_{i+1}$, the color of vertices in $V_{i-1}$ to $c_{i}$, and so on, until changing the color of vertices in $V_{1}$ to $c_{2}$. At this point, $A$ can be colored with $c_{1}$, which contradicts the fact that $H$ is the largest subgraph that can be colored. (15 points)
(2) For any vertex in $V_{k}$, there exists a path such that it can be reached from $A$ by passing through the colors $c_{1} \rightarrow c_{2} \rightarrow \ldots \rightarrow c_{k} \rightarrow c_{1} \rightarrow \ldots \rightarrow c_{k}$ (possibly only one cycle).
This conclusion is guaranteed by the process of defining the sets.
(3) There exists a vertex in $V_{k}$ that is adjacent to $A$.

Otherwise, we change the color of vertices in $V_{k}$ to $c_{1}$, the color of vertices in $V_{k-1}$ to $c_{k}$, and so on, until changing the color of vertices in $V_{1}$ to $c_{2}$. At this point, $A$ can be colored with $c_{1}$, which contradicts the fact that $H$ is the largest subgraph that can be colored. (18 points)

Thus, we have shown that for any color sequence $c_{1}, c_{2}, \ldots, c_{k}$, there always exists a cycle starting from $A$ and passing through such a color sequence several times before returning to $A$. These cycles are clearly different, and in particular, if the sequence is empty, it corresponds to a single person's seating arrangement, and if the sequence is one color, it corresponds to two people's seating arrangement. Therefore, the number of valid seating arrangements is no less than the number of such color sequences, i.e.,
$$
1+n+n(n-1)+n(n-1)(n-2)+\ldots+n!+n!=n!\sum_{i=0}^{n} \frac{1}{i}=[e \cdot n!] \text {. (21 points) }
$$