V OBM

Suppose without loss of generality that $a\le b\le c$. We must have $a\le 3\cdot 1983$, 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 $b\le \frac{2}{k}$, 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.