I will answer your second question first : If $$\frac1x + \frac1y = \frac1{n}$$ For some nonzero positive integer $n$ then it's clear by rearranging that $n(x + y) = xy$. If $(x, y) = k$, i.e., $x = a \cdot k$ and $y = b \cdot k$ where $(a, b) = 1$, then $n \cdot (a + b) = k \cdot ab$. Thus, $a | n$. But then $n/a \cdot ( a + b) = k \cdot b$, and $b$ must divide $n/a$ (which is an integer). Thus, the number of solutions $(a, b)$ are
$$\sum_{a | n} \sum_{b | n/a} 1 = \tau(n)$$
Where $\tau(n)$ is the number of divisors of $n$. In this case, however $1987$ is a prime, thus $\tau(1987) = 2$. This makes answering the former question a lot easier : one clear solution is $(3974, 3974)$ as $1/2 + 1/2 = 1$. Another solution is $(1988, 3950156)$, coming from a little nonobvious identity $(a + 1)^{-1} + \left ( a \cdot (a + 1) \right )^{-1} = a^{-1}$. By the calculations above, these are all the possible solutions. It is thus clear that $\text{max}(x + y) = 1988 + 3950156 = 3952144$ and $\text{min}(x) = 1988$.
