Solution: The greatest common divisor of $a$ and $b$ must divide their sum $a + b = 100$. The largest divisor of 100 is 50. To achieve $\gcd(a, b) = 50$, set $a = 50$ and $b = 50$. Thus, the maximum value is $oxed{50}$.

["The Greatest Common Divisor of Two Numbers Must Divide Their Sum\nUnderstanding Why gcd(a, b) | (a + b) and How to Maximize It", "When working with two integers (a) and (b), one fundamental property of the greatest common divisor (gcd) plays a crucial role in number theory: the gcd of (a) and (b) must divide their sum (a + b). This principle is both elegant and powerful, offering insight into how divisors interact across pairs of numbers.", "### Why Does the GCD Divide the Sum?", "Let (d = \gcd(a, b)). By definition, (d) divides both (a) and (b). This means we can write:", "[\na = d \cdot m \quad \ ext{and} \quad b = d \cdot n\n]", "for some integers (m) and (n) with (\gcd(m, n) = 1). Adding these gives:", "[\na + b = d(m + n)\n]", "This shows that (d) divides (a + b), confirming the core rule: (\gcd(a, b)) divides (a + b).", "### Applying the Rule to (a + b = 100)", "Given (a + b = 100), since (d = \gcd(a, b)) must divide 100, the possible values of (d) are the positive divisors of 100:", "[\n1, 2, 4, 5, 10, 20, 25, 50, 100\n]", "But not every divisor is achievable — only those (d) for which (a = d \cdot m) and (b = d \cdot n) exist with (m + n = \frac{100}{d}) and (\gcd(m, n) = 1).", "To maximize (\gcd(a, b)), we seek the largest divisor (d) such that two integers (m) and (n) exist satisfying:", "[\nm + n = \frac{100}{d} \quad \ ext{and} \quad \gcd(m, n) = 1\n]", "Since (a = 50) and (b = 50) gives:", "[\na + b = 100, \quad \gcd(50, 50) = 50, \quad \ ext{and} \quad 50 \mid 100\n]", "this configuration is valid and achieves (\gcd = 50), the maximum possible divisor of 100.", "### Why 50 is the Largest Possible", "Suppose (\gcd(a, b) = d > 50). The next divisors are 100. If (d = 100), then (a + b = 100) implies (a = 100), (b = 0), but (b = 0) is not valid in this context (gcd(100, 0) is 100, but traditionally (\gcd(a, 0) = |a|), so technically allowed — however, the problem typically considers positive integers). Even if allowed, (b = 0) is often excluded in gcd contexts.", "Thus, the largest feasible (\gcd) with positive integers summing to 100 is indeed 50, achieved when (a = b = 50).", "### Final Answer", "The greatest common divisor of two positive integers (a) and (b) must divide their sum (a + b). For (a + b = 100), the largest possible value of (\gcd(a, b)) is ( \boxed{50} ).", "Set (a = 50), (b = 50) — their gcd is 50, and 50 divides 100, confirming the pattern. This example illustrates how divisor constraints shape the possible shared factors of integer pairs."]