Romanian Mathematical Olympiad

Since the fractions $f_1 = \frac{a+1}{b}$ and $f_2 = \frac{b+2}{a}$ are positive integers, $a+1 \ge b$ and $b+2 \ge a$, hence $a+1 \ge b \ge a-2$. This leaves the cases:

1. $b = a + 1$: then $f_2 = 1 + \frac{3}{a} \in \mathbb{N}$, whence $a = 1, b = 2$ or $a = 3, b = 4$, which are convenient values.

2. $b = a$: then $f_1 = 1 + \frac{1}{a} \in \mathbb{N}$, whence $a = 1, b = 1$, which are convenient values.

3. $b = a - 1$: then $f_2 = 1 + \frac{1}{a} \in \mathbb{N}$, whence $a = 1, b = 1$.

4. $b = a - 2$: then $f_1 = 1 + \frac{3}{b} \in \mathbb{N}$, whence $a = 3, b = 1$ or $a = 5, b = 3$, which are convenient values.