Solution: It is evident from the problem statement that we can find 1987 sets, all of which contain a common element $a$, and any two sets do not have a common element other than $a$.
However, is this the only possibility?
Given that the union of any two sets has 89 elements, it is known that any two of the 1987 sets have exactly one common element. It is easy to prove that there must be an element $a$ in one of these 1987 sets that appears in 45 other sets outside of $A$, denoted as $A_{1}, A_{2}, \cdots, A_{46}$, and the rest are denoted as $A_{46}, A_{47}, \cdots, A_{1986}$.

Let $B$ be any set among $A_{46}, \cdots, A_{1986}$, and $a \notin B$. By the problem statement, $B$ and $A, A_{1}, A_{2}, \cdots, A_{45}$ each have one common element, and these 46 elements are all different, so $B$ must have 46 elements, which contradicts the problem statement. Therefore, all 1987 sets contain $a$.

Thus, the result is $1987 \times 44 + 1 = 87429$, meaning the union of these 1987 sets has 87429 elements.
Example 8 has 1987 sets, each set has 45 elements, and the union of any two sets has 89 elements. How many elements are there in the union of these 1987 sets.