Assume integer $m \geq 2.$ There are $3m$ people in a meeting, any two of them either shake hands with each other once or not.We call the meeting "$n$-interesting", only if there exists $n(n\leq 3m-1)$ people of them, the time everyone of whom shakes hands with other $3m-1$ people is exactly $1,2,\cdots,n,$ respectively. If in any "$n$-interesting" meeting, there exists $3$ people of them who shake hands with each other, find the minimum value of $n.$
To solve this problem, we need to find the minimum value of \( n \) such that in any \( n \)-interesting meeting, there exists a set of 3 people who all shake hands with each other. Let's break down the solution step by step.

1. **Constructing a \( 2m \)-interesting meeting:**

   We partition the \( 3m \) people into three groups \( A, B, \) and \( C \) with \( m \) people in each group. Let's denote the people in group \( A \) as \( A_1, A_2, \ldots, A_m \), in group \( B \) as \( B_1, B_2, \ldots, B_m \), and in group \( C \) as \( C_1, C_2, \ldots, C_m \).

   - Each person \( B_i \) (where \( 1 \leq i \leq m \)) will shake hands with \( A_1, A_2, \ldots, A_i \) and all \( m \) people in group \( C \).

   - Therefore, the degrees of people in group \( A \) will be \( 1, 2, \ldots, m \) respectively, and the degrees of people in group \( B \) will be \( m+1, m+2, \ldots, 2m \) respectively.

   - This construction ensures that the \( 2m \) people in groups \( A \) and \( B \) form a \( 2m \)-interesting meeting, and there are no triangles (sets of 3 people who all shake hands with each other).

2. **Proving that \( 2m+1 \) is the minimum value:**

   Suppose we have a \( 2m+1 \)-interesting meeting. Consider the person with degree \( 2m+1 \), call this person \( X \).

   - Since \( X \) shakes hands with \( 2m+1 \) people, there are \( m-1 \) people who do not shake hands with \( X \).

   - Among the \( 2m+1 \) people who shake hands with \( X \), there must be at least \( m \) people with degrees greater than \( m \) (since the degrees range from \( 1 \) to \( 2m+1 \)).

   - Let \( Y \) be one of these people with degree greater than \( m \) who shakes hands with \( X \). Since \( Y \) has a degree greater than \( m \), \( Y \) must shake hands with another person \( Z \) among the \( 2m+1 \) people who also shake hands with \( X \).

   - Therefore, \( X, Y, \) and \( Z \) form a triangle, as they all shake hands with each other.

This proves that in any \( 2m+1 \)-interesting meeting, there exists a set of 3 people who all shake hands with each other. Hence, the minimum value of \( n \) is \( 2m+1 \).

The final answer is \( \boxed{ 2m+1 } \).