Because $$\frac{a}{b} + \frac{14b}{9a}=\frac{9a^2 + 14b^2}{9ab}\implies 9ab\mid 9a^2 + 14b^2$$

It is clear that $b$ must be a multiple of $3$. However, $b$ cannot divide $9$ otherwise both $9ab$ and $14b^2$ are multiple of $81$. This will force $a$ to be a multiple of $3$. This contradicts the claim that $a$ and $b$ are co-prime. Therefore, we conclude $b = 3$. It follows that $$9ab\mid 9a^2 + 14b^2\implies 9\times a\times 3\mid 9a^2 + 14\times 9 \implies 3a\mid a^2 +14\implies a\mid 14$$

Hence, $a \in \{1,2,7,14\}$. It can be verified that all these $\boxed{4}$ pairs satisfy the requirement.