Easy peasy

Let's rationalize this expression first so it's easier to deal with: $\frac{\sqrt{2} + \sqrt{a}}{\sqrt{3} + \sqrt{b}} = \frac{(\sqrt{2} + \sqrt{a})(\sqrt{3} - \sqrt{b})}{(\sqrt{3} + \sqrt{b})(\sqrt{3} - \sqrt{b})} = \frac{ \sqrt{6} - \sqrt{2b} + \sqrt{3a} - \sqrt{ab} } {3-b}$

Straight away we see that $b$ can't be $3$ . Since this number has to be rational, the only irrational term we see so far is $\sqrt{6}$ , so one of the other negative radicals has to be $\sqrt{6}$ to cancel it; either $\sqrt{2b}$ or $\sqrt{ab}$ .

$\sqrt{2b}$ obviously won't work as $b$ can't be $3$ , so $\sqrt{6} = \sqrt{ab}$ . Now we look for positive-integer pairs that multiply to $6$ .

$(a, b)$ can be either $(3, 2)$ , $(6, 1)$ , or $(1, 6)$ . The latter two don't give a rational number when substituted into the equation, so $(a, b) = (3, 2)$ is the only solution.

$a + b = 3 + 2 = \boxed{5}.$

Can be solved by simple logic. The most basic values of a and b for it to be a rational number are 3 and 2 as the numerator and the denominator get canceled then and the value will be one which is rational.