Brazilian Mathematical Olympiad

Suppose without loss of generality that $a \leq b \leq c$. We must have $a \leq 3 \cdot 1983$, otherwise $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} < \frac{1}{1983}$. So there are only finitely many possible values for $a$.

Now consider the number of solutions for fixed $a$. We have $\frac{1}{b} + \frac{1}{c} = \frac{1}{1983} - \frac{1}{a} = k$. Now we must have $b \leq \frac{2}{k}$, otherwise $\frac{1}{b} + \frac{1}{c} < k$. So there are only finitely many possible values for $b$. The number $c$ is fixed once $a$ and $b$ are fixed, so we have shown that there are only finitely many solutions.