Let $u=\frac{a}{b}.$ Then the problem is equivalent to finding all possible rational numbers 'u' such that
$u+\frac{14}{9u}=K$, for some integer K.
This equation is equivalent to $9u^2-9uk+14=0$
$9u^2-9uk+14=0$ whose solutions are,
$u=\frac{9k\pm \sqrt{81k^2-504}}{18}$
$=\frac{K}{2}\pm \frac{1}{6}\sqrt{9k^2-56}$
Hence u is rational if and onlly if
$\sqrt{9k^2-56}$ is rational, Which is true if and only if $9k^2-56$ is a perfect square. Suppose that $9k^2-56=S^2$ for some positive integer S.
(3k+s)(3k-S)=56
The only factors of 56 are 1,2,4,7,8,14,28 and 56. So, (3k-S) and (3k+s) is one of the
ordered pairs (1,56), (2,28), (4,14) and (7,8). The cases (1,56) and (7,8) yield no (4,14) yield k=5 and k=3 respectively.
If k=5 then $u=\frac{1}{3}$ or $u=\frac{14}{3}$
If k=3 then $u=\frac{2}{3}$ of $u=\frac{7}{3}.$
Therefore there are four pairs (a,b) that satisfy the given conditions, namely (1,3), (2,3), (7,3) and (14,3)