Solution:

Combining, the expression is $\dfrac{9a^2+14b^2}{9ab}.$ For this to be an integer, $a$ must divide $9a^2+14b^2,$ hence $a\mid 14b^2.$ Since $\gcd(a,b)=1,$ we get $a\mid 14.$ Similarly $b\mid 9a^2$ gives $b\mid 9.$

So $a\in\{1,2,7,14\}$ and $b\in\{1,3,9\}.$ Checking these, only $b=3$ makes the expression an integer for each allowed $a.$

The valid pairs are $(1,3),(2,3),(7,3),$ and $(14,3),$ for a total of $4.$

Thus, the correct answer is A.