Example 1. A newly built city has purchased a large number of public buses to solve the city's public transportation problems. They plan to set up bus stops at 1983 different locations and connect them by opening several bus routes. Their wishes are:
(i) to open as many routes as possible,
(ii) every two routes should have at least one common bus stop;
(iii) each bus stop should be shared by at most two different routes. How many bus routes can they open at most according to these wishes? How many bus stops should each route have at least?
Let a total of $k$ bus routes be established, and let the set of 1983 bus stops be denoted as $\mathrm{A}$. The sets of bus stops served by each route are respectively denoted as
$$
A_{1}, \hat{H}_{2}, \cdots, \hat{A}_{K},
$$

It can be known from the problem statement that these subsets have the following properties:
(i) $A_{i_{1}} \cap A_{i_{2}} \neq \phi, i_{1} \neq i_{2}$;
(ii) $A_{i_{1}} \cap A_{i_{2}} \cap A_{i_{3}} = \phi, i_{1}, i_{2}, i_{3}$ are pairwise distinct. Therefore, each subset $A_{i}$ contains at least $k-1$ elements, because by (i) it has at least one common element with each of the other subsets, and by (ii) these elements are pairwise distinct. Thus, the sum of the number of elements in these $k$ subsets is at least $k(k-1)$.

However, in the above calculation, each element is counted twice, so the number of elements in $\bigcup_{i=1}^{k} A_{i}$ is at least $\frac{k(k-1)}{2}$. But we should have $\bigcup_{i=1}^{k} A_{i} = A$, so $\frac{k(k-1)}{2} \leq 1983$.

The largest integer satisfying this inequality is $k=63$, so they can establish at most 63 bus routes, and each route should serve at least 62 bus stops.