Suppose \( n = \frac{a}{b} + \frac{a+1}{b+1} \), then \[ n - 2 = \frac{(a-b)(2b+1)}{b(b+1)}. \] Since \(\gcd(2b+1, b) = \gcd(2b+1, b+1) = 1\), it follows that \( 2b+1 \mid (n-2) \). This implies that \(|n-2|\) cannot be a power of 2, i.e., \( n \) cannot be \( 1 \) or \( 2^k + 2 \) for any \( k \geq 0 \). On the other hand, suppose \( n-2 \) has a positive divisor \( 2b+1 \geq 3 \). Then, let \[ a = \frac{n-2}{2b+1} \cdot b(b+1) + b. \] It is evident that this choice of \( a \) satisfies the given condition. Therefore, all positive integers \( n \) are of the form \( n = 1 \) or \( n = 2^k + 2 \) for \( k \geq 0 \).