imc

Let $\frac{a}{b}=r.$ So $r+\frac{14}{9r}=I$, where I is an integer. Algebraic manipulations yield: $r^2-Ir+\frac{14}{9}=0$. The discriminant of this must be the square of a rational number, call this R. So $I^2-\frac{56}{9}=R^2⟶I^2-R^2=(I-R)(I+R)=\frac{56}{9}$. I is $\frac{1}{2}$ the sum of $I-R$ and $I+R$. To have an integer sum, $I-R$ and $I+R$ must have the same denominator, namely 3. We proceed with casework. Case 1. $I+R=56/3$, $I-R=1/3$. This yields $I=19/2$, which is not an integer. This case produces 0 solutions. Case 2. $I+R=28/3$, $I-R=2/3$. This yields $I=5$. Substituting into our original equation yields: $r^2-5r+\frac{14}{9}=0$. Factoring gives: $r=\frac{1}{3}$, $r=\frac{14}{3}$. This case produces 2 solutions, namely (1,3) and (14,3). Case 3. $I+R=14/3$, $I-R=4/3$. This yields $I=3$. Substituting into our original equation yields: $r^2-3r+\frac{14}{9}=0$. Factoring gives: $r=\frac{2}{3}$, $r=\frac{7}{3}$. This case produces 2 solutions, namely (2,3) and (7,3). Case 4. $I+R=8/3$, $I-R=7/3$. This yields $I=5/2$, which is not an integer. This case produces 0 solutions. Altogether, we have 4 solutions, so our answer is $\boxed{4}$.