Sكية: The greatest common divisor of $a$ and $b$ must divide their sum, which is 2025. To maximize $\gcd(a,b)$, we find the largest proper divisor of 2025. Factoring 2025: $2025 = 3^4 \times 5^2$. The largest proper divisor is $2025 / 3 = 675$. Thus, the maximum $\gcd(a,b)$ is $\boxed{675}$.

["Title: Unlocking the Secrets of Greatest Common Divisor: When gcd(a, b) Divides a + b = 2025", "Ever wondered why the greatest common divisor (gcd) of two integers always divides their sum? The fascinating link between gcd and numerical relationships gives us an elegant rule, especially when that sum is fixed—like 2025. This article explores why $\gcd(a, b)$ must divide $a + b$, how to maximize it, and unveils the secret behind the maximum $\gcd$ when $a + b = 2025$.", "## The Fundamental Rule: gcd(a, b) Divides a + b", "One of the foundational insights in number theory is that the greatest common divisor of two integers $a$ and $b$ always divides their sum:\n$$\n\gcd(a, b) \mid (a + b)\n$$\nThis means that $\gcd(a, b)$ must be a divisor of the total sum. When $a + b = 2025$, any valid $\gcd(a, b)$ must therefore be one of the divisors of 2025.", "## Why Does the gcd Divide the Sum?", "Set $d = \gcd(a, b)$. Then $a = d \cdot m$ and $b = d \cdot n$ for some integers $m, n$ with $\gcd(m, n) = 1$. The sum becomes:\n$$\na + b = d(m + n) = d \cdot k \quad \ ext{for some integer } k\n$$\nSo $d$ divides $a + b$ — a direct consequence of scalar multiplication and divisibility.", "## Maximizing gcd(a, b) When a + b = 2025", "To maximize $\gcd(a, b)$, we seek the largest proper divisor of 2025 — because using the full 2025 would mean $a = b = 2025$, which gives $a + b = 4050$, not 2025.", "We begin by factoring 2025:\n$$\n2025 = 3^4 \ imes 5^2\n$$\nThis composite structure reveals all divisors of 2025. The largest proper divisor is obtained by removing the smallest prime factor repeated most: simply divide 2025 by its smallest prime factor, which is 3:\n$$\n\frac{2025}{3} = 675\n$$", "Thus, the maximum possible $\gcd(a, b)$ when $a + b = 2025$ is 675.", "## Verifying: Can gcd(a, b) = 675?", "Let $a = 675m$, $b = 675n$, with $\gcd(m, n) = 1$ and $m + n = \frac{2025}{675} = 3$. Choose coprime $m, n$ such that their sum is 3 — e.g., $m = 1$, $n = 2$. Then:\n$$\na = 675 \ imes 1 = 675,\quad b = 675 \ imes 2 = 1350\n$$\nClearly, $\gcd(675, 1350) = 675$, and $a + b = 2025$, confirming our result.", "## Final Answer", "Therefore, the largest possible value of $\gcd(a, b)$ when $a + b = 2025$ is:\n$$\n\boxed{675}\n$$", "---", "### Key Takeaways", "- The gcd of two numbers always divides their sum.\n- Maximizing gcd under a fixed sum means finding the largest proper divisor of that sum.\n- Factor wisely and reduce using prime factors to uncover the largest valid divisor.\n- This insight unlocks efficient computation in number theory and competitive math.", "Try it yourself: Pick another sum, factor it, and find its largest proper divisor — your gcd can go no higher!"]