Four. (50 points) Given a positive integer $n(n \geqslant 2)$. From 1, $2, \cdots, 3 n$, any $m$ numbers are chosen such that there must be four pairwise distinct numbers $a$, $b$, $c$, $d$, such that $a=b+c+d$. Find the minimum value of $m$.

保留源文本的换行和格式，直接输出翻译结果如下：

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Four. (50 points) Given a positive integer $n(n \geqslant 2)$. From 1, $2, \cdots, 3 n$, any $m$ numbers are chosen such that there must be four pairwise distinct numbers $a$, $b$, $c$, $d$, such that $a=b+c+d$. Find the minimum value of $m$.
```
Four, $m_{\min }=2 n+2$.
Take $n, n+1, \cdots, 3 n$ these $2 n+1$ numbers.
Since the sum of any three numbers is greater than $3 n$, hence
$$
m_{\text {min }}>2 n+1 \text {. }
$$

Now consider taking any $2 n+2$ numbers from $1,2, \cdots, 3 n$, denoted as $a_{1}<a_{2}<\cdots<a_{2 n+2}$.
Notice,
$$
\begin{array}{l}
a_{2 n+2}-a_{2}, a_{2 n+2}-a_{3}, \cdots, a_{2 n+2}-a_{2 n+1}, \\
a_{1}+a_{2}, a_{1}+a_{3}, \cdots, a_{1}+a_{2 n+1}
\end{array}
$$

There are a total of $4 n$ numbers, all within the interval $[3,4 n-2]$.
Thus, there must be two pairs of numbers that are equal, i.e.,
$$
\begin{array}{l}
a_{2 n+2}-a_{i}=a_{1}+a_{j}, a_{2 n+2}-a_{k}=a_{1}+a_{l} . \\
\text { If } i=j, k=l \text {, then } \\
2 a_{i}=a_{2 n+2}-a_{1}=2 a_{k} \Rightarrow i=k,
\end{array}
$$

which is a contradiction.
Therefore, there must be $i \neq j$ or $k \neq l$.