Example 4 Given 2003 sets, each set contains exactly 44 elements, and any two sets have exactly one common element. Try to find the number of elements in the union of these 2003 sets.
Solution 1 Let the given 2003 sets be $A_{1}, A_{2}, \cdots, A_{2003}$. According to the problem, we have $\left|A_{i}\right|=44(1 \leqslant i \leqslant 2003)$. $\left|A_{i} \cap A_{j}\right|=1(1 \leqslant i<j \leqslant 2003)$. To find $\left|A_{1} \cup A_{2} \cup \cdots \cup A_{2003}\right|$, by the principle of inclusion-exclusion, we only need to find $\left|A_{i_{1}} \cap A_{i_{2}} \cap \cdots \cap A_{i_{k}}\right|\left(1 \leqslant i_{1}<i_{2}<\cdots<i_{k} \leqslant 2003, k \geqslant 3\right)$. From $\left|A_{i} \cap A_{j}\right|=1$ we know this number is at most 1, and if all are equal to 1, then there must be an element that is a common element of all $A_{i}$.

Consider $A_{1}$. It has a unique common element with each of the other 2002 sets $A_{i}(2 \leqslant i \leqslant 2003)$. Also, $\left|A_{1}\right|=44$, and $\left[\frac{2003}{44}\right]=45$. If each element of $A_{1}$ is in at most 45 other sets, then $A_{1}$ can have a common element with at most $44 \times 45=1980$ sets, which is a contradiction. Therefore, $A_{1}$ must have an element $a$ that is in at least 46 other sets. Without loss of generality, assume $a$ is in $A_{2}, A_{3}, \cdots, A_{47}$, and let $B$ be any of the sets $A_{48}, A_{49}, \cdots, A_{2003}$. If $a \notin B$, and since $B$ has a common element with each of $A_{1}, A_{2}, \cdots, A_{47}$, and these common elements are all different (otherwise, if $B$ has the same common element $b$ with $A_{i}$ and $A_{j}$ $(1 \leqslant i<j \leqslant 47)$, then $b \neq a$, and $A_{i}$ and $A_{j}$ would have two common elements $a$ and $b$, which is a contradiction), then $B$ must have at least 47 elements, which contradicts $|B|=44$. Therefore, $a \in B$, meaning $a$ is a common element of the 2003 sets $A_{1}, A_{2}, \cdots, A_{2003}$. But any two of $A_{1}, A_{2}, \cdots, A_{2003}$ have only one common element $a$, so
$$
\left|A_{i_{1}} \cap A_{i_{2}} \cap \cdots \cap A_{i_{k}}\right|=1\left(1 \leqslant i_{1}<\cdots<i_{k} \leqslant 2003\right) .
$$

By the principle of inclusion-exclusion, we get
$$
\begin{aligned}
& \left|A_{1} \cup A_{2} \cup \cdots \cup A_{2003}\right| \\
= & \sum_{i=1}^{203}\left|A_{i}\right|-\sum_{1 \leqslant i<j \leqslant 20033}\left|A_{i} \cap A_{j}\right|+\sum_{1 \leqslant i<j<k \leqslant 2003}\left|A_{i} \cap A_{j} \cap A_{k}\right|-\cdots+\left|A_{1} \cap A_{2} \cap \cdots \cap A_{2003}\right| \\
= & 2003 \times 44-C_{2003}^{2}+C_{2003}^{3}-\cdots+C_{2003}^{2003} \\
= & 2003 \times 44+C_{2003}^{0}-C_{2003}^{1}-\left(C_{2003}^{0}-C_{2003}^{1}+C_{2003}^{2}-\cdots C_{2003}^{2003}\right) \\
= & 86130-(1-1)^{2003}=86130 .
\end{aligned}
$$

Solution 2 From the same reasoning as in Solution 1, we know that $a$ is a common element of the 2003 sets $A_{1}, A_{2}, \cdots, A_{2003}$, and any two of $A_{1}, A_{2}, \cdots, A_{2003}$ have only one common element $a$. Each set has 44 elements, and the 2003 sets have a total of $2003 \times 44$ elements. However, only the element $a$ is counted 2003 times. Therefore,
$$
\left|A_{1} \cup A_{2} \cup \cdots \cup A_{2003}\right|=2003 \times 44-(2003-1)=86130 .
$$