6. Given an integer $m \geqslant 2$. A meeting has a total of $3 m$ representatives attending, and any two representatives either shake hands once or do not shake hands. For a positive integer $n (n \leqslant 3 m-1)$, if there exist $n$ representatives among them whose number of handshakes are $1, 2, \cdots, n$, then the meeting is called “$n$-interesting”. If for all possible $n$-interesting meetings, there always exist three representatives who have all shaken hands with each other, find the minimum value of $n$.
(Zhang Pengcheng, problem contributor)
6. Let the $3m$ representatives be denoted as $A_{1}, A_{2}, \cdots, A_{3 m}$.

When $1 \leqslant n \leqslant 2 m$, construct an $n$-interesting meeting as shown in Figure 7: for $i=1,2, \cdots, m$, let $A_{3 m+1-i}$ shake hands with $A_{i}, A_{i+1}, \cdots, A_{2 m}$.
At this point, the number of handshakes for representatives $A_{1}, A_{2}, \cdots, A_{m}, A_{2 m+1}, A_{2 m+2}, \cdots, A_{3 m}$ are $1,2, \cdots, 2 m$ respectively.

Notice that $2 m \geqslant n$, this is an $n$-interesting meeting. For any three representatives, there must be two representatives who both belong to $S_{1}=\left\{A_{1}, A_{2}, \cdots, A_{2 m}\right\}$ or both belong to $S_{2}=\left\{A_{2 m+1}, A_{2 m+2}, \cdots, A_{3 m}\right\}$. By the construction of the graph, these two representatives have not shaken hands. Therefore, there do not exist three representatives who have all shaken hands with each other.
Thus, $n \geqslant 2 m+1$.
When $n=2 m+1$, consider any $(2 m+1)$-interesting meeting. In this meeting, there exists one representative who has shaken hands exactly $2 m+1$ times. Without loss of generality, assume $A_{3 m}$ has shaken hands with $A_{1}, A_{2}, \cdots, A_{2 m+1}$, these $2 m+1$ representatives.
$$
\begin{array}{l}
\text { Let } T_{1}=\left\{A_{1}, A_{2}, \cdots, A_{2 m+1}\right\}, \\
T_{2}=\left\{A_{2 m+2}, A_{2 m+3}, \cdots, A_{3 m}\right\} .
\end{array}
$$

Since the number of handshakes for representatives in $T_{2}$ can take at most $m-1$ different values from $\{m, m+1, \cdots, 2 m+1\}$, there exists a number $k \in\{m, m+1, \cdots, 2 m+1\}$ that is not the number of handshakes for any representative in $T_{2}$. This indicates that there must be a representative in $T_{1}$ whose number of handshakes is $k$, let this representative be $A_{i}$. Since $A_{i}$ has shaken hands with at most $m-1$ representatives in $T_{2}$, there must be a representative $A_{j}\left(A_{j} \neq A_{i}\right)$ in $T_{1}$ who has shaken hands with $A_{i}$, thus $A_{3 m}, A_{i}, A_{j}$ have all shaken hands with each other.

Therefore, in any $(2 m+1)$-interesting meeting, there always exist three representatives who have all shaken hands with each other.
In summary, the minimum value of $n$ is $2 m+1$.