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📰 Thus, \(d\) must be a divisor of 2024, and \(m + n = \frac{2024}{d}\). To maximize \(d\), we minimize \(m + n\), which is minimized when \(m + n = 2\), since \(m, n \geq 1\) and coprime. The only such pair is \(m = 1\), \(n = 1\), which are coprime. Then:
📰 d = \frac{2024}{2} = 1012
📰 This is valid because \(a = 1012\), \(b = 1012\), and \(\gcd(1012, 1012) = 1012\). Therefore, the maximum possible value is \(\boxed{1012}\).
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