Question 224, Given a positive integer $n(n \geq 2)$, choose $m$ different numbers from $1, 2, \ldots, 3n$. Among these, there must be four pairwise distinct numbers $a, b, c, d$, satisfying $a=b+c+d$. Find the minimum value of $m$.

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Question 224, Solution: On the one hand, take the numbers $n, n+1, \ldots, 3n$, which are $2n+1$ numbers. Since the sum of any three of these numbers is greater than $3n$, there do not exist four distinct positive integers $a, b, c, d$ such that $a=b+c+d$. Therefore, $m_{\text{min}} > 2n+1$.

Next, consider selecting $2n+2$ numbers from $1, 2, \ldots, 3n$, denoted as $a_1 < a_2 < \ldots < a_{2n+2}$. Note that the numbers $a_{2n+2} - a_{2n+1}, a_{2n+2} - a_{2n}, \ldots, a_{2n+2} - a_2$ and $a_1 + a_2, a_1 + a_3, \ldots, a_1 + a_{2n+1}$, a total of $4n$ numbers, all fall within the interval $[1, 4n-2]$. Thus, there must be two pairs of numbers that are equal, say $a_{2n+2} - a_i = a_1 + a_j$ and $a_{2n+2} - a_s = a_1 + a_t$. If $i = j$ and $s = t$, then $2a_i = 2a_s = a_{2n+2} - a_1$, which implies $i = s$, leading to a contradiction.
In conclusion, the minimum value of $m$ that satisfies the condition is $m_{\min} = 2n+2$.