Solution: Let $a + b = 2025$. The greatest common divisor $\gcd(a, b)$ must divide $a + b = 2025$. The maximum possible value of $\gcd(a, b)$ occurs when $a$ and $b$ are both multiples of the largest divisor of 2025. Since $2025 = 45^2 = 3^4 \cdot 5^2$, its largest proper divisor is $2025 / 3 = 675$. Thus, $\gcd(a, b) = 675$ when $a = 675$ and $b = 1350$.

["Title: Maximize $\gcd(a, b)$ Given $a + b = 2025$ — The Key Lies in Divisors of 2025", "Meta Description:\nExplore why the greatest common divisor $\gcd(a, b)$ of two positive integers $a$ and $b$ summing to 2025 must divide 2025. Learn the maximum possible value of $\gcd(a, b)$ and how choosing $a = 675$, $b = 1350$ achieves it.", "---", "### Understanding the GCD Constraint When $a + b = 2025$", "Mathematicians often encounter elegant problems where divisibility and greatest common divisors reveal deep structural insights. One such elegant setup is when two positive integers $a$ and $b$ satisfy:", "$$\na + b = 2025\n$$", "Then, a fundamental number theory principle tells us:\nThe greatest common divisor $\gcd(a, b)$ must divide their sum $a + b = 2025$.", "This property arises because $\gcd(a, b)$ divides both $a$ and $b$, so it must also divide any linear combination — including $a + b$. Therefore:", "$$\n\gcd(a, b) \mid 2025\n$$", "This simple insight opens the door to finding the maximum possible value of $\gcd(a, b)$.", "---", "### Finding the Maximum Possible $\gcd(a, b)$", "Let $d = \gcd(a, b)$. Then we can write:", "$$\na = d \cdot m, \quad b = d \cdot n\n$$", "where $m$ and $n$ are coprime integers ($\gcd(m, n) = 1$). Since $a + b = 2025$, substituting gives:", "$$\nd(m + n) = 2025 \quad \Rightarrow \quad d \mid 2025 \ ext{ and } m + n = \frac{2025}{d}\n$$", "For $d$ to be as large as possible, the value $\frac{2025}{d}$ must still allow $m$ and $n$ to be positive coprime integers summing to that quotient.", "The largest possible $d$ occurs when $\frac{2025}{d}$ is minimized but still enables valid $m, n$. But since $m$ and $n$ must be at least 1 (because $a, b > 0$), the smallest allowable $m+n$ is 2. However, to maximize $d$, we instead consider the largest divisor $d$ such that $m + n = \frac{2025}{d}$ has at least one coprime pair $(m, n)$.", "But the best scenario — achieving the maximum $d$ — happens when:", "$$\nm + n = \frac{2025}{d} \geq 2 \quad \ ext{and} \quad \gcd(m, n) = 1\n$$", "However, the absolute maximum value of $d$ occurs not just from divisibility but from how perfectly $d$ divides 2025 and how its cofactors behave.", "Since $2025 = 3^4 \cdot 5^2$, its total number of positive divisors is:", "$$\n(4+1)(2+1) = 15\n$$", "We aim to find the largest divisor $d$ such that there exist positive integers $m, n$ with $m + n = \frac{2025}{d}$ and $\gcd(m, n) = 1$.", "But here’s the key: if $d$ divides 2025, then $a = dm$, $b = dn$, and $\gcd(a, b) = d$. So the maximum $\gcd(a, b)$ must be a divisor of 2025.", "Now, consider the divisor $675$. Note:", "$$\n2025 \div 675 = 3\n$$", "So $m + n = 3$, and we can choose $m = 1$, $n = 2$ (or vice versa). These are coprime, so $\gcd(1,2) = 1$, satisfying all conditions.", "Thus:", "$$\na = 675 \cdot 1 = 675, \quad b = 675 \cdot 2 = 1350\n$$", "$$\n\gcd(675, 1350) = 675, \quad \ ext{and } 675 + 1350 = 2025\n$$", "This confirms that $\gcd(a, b) = 675$ is achievable.", "---", "### Why 675 Is the Maximum", "Suppose $\gcd(a, b) = d > 675$. Then $d$ must divide 2025, so possible candidates are $d = 2025$, $d = 1350$, $d = 675$, etc.", "- $d = 2025$: Then $a = 2025$, $b = 0$, but $b$ must be positive → invalid.\n- $d = 1350$: Then $a + b = 1350(m + n) = 2025 \Rightarrow m + n = \frac{2025}{1350} = 1.5$, not an integer → impossible.\n- $d = 675$: Works perfectly, as shown.", "No larger divisor of 2025 satisfies the requirement that both $a, b > 0$ and $\gcd(a, b) = d$ with $a + b = 2025$ and $m + n = 2025/d$ allowing coprime $m, n$.", "Thus, the maximum possible $\gcd(a, b)$ is 675.", "---", "### Conclusion", "When $a + b = 2025$, the greatest common divisor $\gcd(a, b)$ must divide 2025. By expressing $a = dm$, $b = dn$, with $d = \gcd(a, b)$, we find that $d$ divides 2025 and $m + n = 2025/d$. The largest such $d$ for which $m$ and $n$ are positive integers summing to $2025/d$ and coprime is maximized when $d = 675$, achieved by $a = 675$, $b = 1350$.", "This elegant result highlights how number theory principles simplify complex constraints — all stemming from the simple yet profound truth: $\gcd(a, b)$ divides $a + b$.", "For maximum $\gcd(a, b)$ under $a + b = 2025$, the answer is:", "$$\n\boxed{675}\n$$", "---", "Keywords:\n$\gcd(a,b)$, $a + b = 2025$, maximum gcd, number theory, divisors of 2025, $\gcd$ property, divisibility in integers", "Related reads:\n- All divisors of 2025\n- Properties of greatest common divisors\n- Sum of two integers with fixed gcd", "---", "Discover how mathematical constraints unlock optimal solutions — dive deeper into gcd and number theory today!"]