Number-Theoretic Inequality!

We have $\frac1{q^2} > \left| \frac{p}{q} - \frac23 \right| = \left| \frac{3p-2q}{3q} \right| \ge \frac1{3|q|}$ since the numerator is nonzero. So $|q| < 3$ . Looking at $q = \pm 1, \pm 2$ , we quickly get the six answers $(\pm 1, \pm 1), (0, \pm 1), (\pm 1, \pm 2)$ (where the $\pm$ signs are either both $+$ or both $-$ in each ordered pair). So the answer is $\fbox{6}$ .

If we replaced $\frac23$ by an irrational number, Dirichlet's approximation theorem says that there would be infinitely many choices for $(p,q)$ satisfying the inequality.