Example 10 Arrange all positive integers that are coprime with 105 in ascending order. Try to find the 1000th term of this sequence.
Let $S=\{1,2, \cdots, 105\}, A_{1}=\{x \mid x \in S$ and $3 \mid x\}$,
$A_{2}=\{x \mid x \in S$, and $5 \mid x\}, A_{3}=\left\{x \mid x \in S_{1}\right.$ and $\left.7 \mid x\right\}$, then
$$
|S|=105,\left|A_{1}\right|=35,\left|A_{2}\right|=21,\left|A_{3}\right|=15,\left|A_{1} \cap A_{2}\right|=7 \text {, }
$$
$\left|A_{1} \cap A_{3}\right|=5,\left|A_{2} \cap A_{3}\right|=3,\left|A_{1} \cap A_{2} \cap A_{3}\right|=1$, so, the number of natural numbers not greater than 105 and coprime with 105 is $\left|\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3}\right|=|S|-\left(\left|A_{1}\right|+\right.$ $\left.\left.\left|A_{2}\right|+\left|A_{3}\right|\right)+\left(\left|A_{1} \cap A_{2}\right|+\right) A_{2} \cap A_{3}|+| A_{3} \cap A_{1} \mid\right)-\left(A_{1} \cap A_{2} \cap\right.$ $\left.A_{3}\right)=48$.

Let the sequence of positive integers that are coprime with 105, arranged in ascending order, be $a_{1}$, $a_{2}, a_{3}, \cdots, a_{n}$, then $a_{1}=1, a_{2}=2, a_{3}=4, \cdots, a_{48}=104$.

Let $P=\left\{a_{1}, a_{2}, \cdots, a_{48}\right\}$, on one hand, for $n \geqslant 1$, let $a_{n}=105 k+r$ $(k \geqslant 0 ; 0 \leqslant r \leqslant 104, k, r \in Z)$, since $\left(a_{n}, 105\right)=1$, we have $(r, 105)=1$, hence $r \in$ $P$; on the other hand, for any non-negative integer $k$ and $r \in P$, since $(r, 105)=1$, we have $(105 k+r, 105)=1$. Thus, there must be an $n$ such that $a_{n}=105 k+r$. This indicates that the sequence $\left\{a_{n}\right\}$ consists of and only of numbers of the form $105 k+r(k \geqslant 0, k \in Z, r \in P)$ arranged in ascending order.

Since the sequence is increasing, and for each fixed non-negative integer $k$, when $r$ takes all values in $P$, a total of 48 numbers are obtained, and $1000=48 \times 20+40$, hence $a_{1000}=105 \times 20+$ $a_{40}$.
$$
\begin{array}{c}
\text { Also, } a_{48}=104, a_{47}=103, a_{46}=101, a_{45}=97, a_{44}=94, a_{43}=92, a_{42} \\
=89, a_{41}=88, a_{40}=86, \text { so, } a_{1000}=105 \times 20+86=2186 .
\end{array}
$$