Example 3 Arrange the positive integers that are coprime with 105 in ascending order, and find the 1000th term of this sequence.
Solution 1 (Estimation Method) Let the 1000th term of this sequence be $n$. Since $105=3 \times 5 \times 7$, according to the problem, we know that among the positive integers less than or equal to $n$, there are exactly 1000 numbers that are not divisible by any of $3, 5, 7$. Let $S=\{1,2, \cdots, n\}, A_{i}=\{m \mid m \in S$ and $m$ is divisible by $i\} (i=3,5,7)$. Thus, by the principle of inclusion-exclusion, we have
$$
\begin{aligned}
1000 & =\left|\bar{A}_{3} \cap \bar{A}_{5} \cap \bar{A}_{7}\right| \\
& =|S|-\left(\left|A_{3}\right|+\left|A_{5}\right|+\left|A_{7}\right|\right)+\left(\left|A_{3} \cap A_{5}\right|+\left|A_{3} \cap A_{7}\right|+\left|A_{5} \cap A_{7}\right|\right)-\left|A_{3} \cap A_{5} \cap A_{7}\right| \\
& =n-\left(\left[\frac{n}{3}\right]+\left[\frac{n}{5}\right]+\left[\frac{n}{7}\right]\right)+\left(\left[\frac{n}{3 \times 5}\right]+\left[\frac{n}{3 \times 7}\right]+\left[\frac{n}{5 \times 7}\right]-\left[\frac{n}{3 \times 5 \times 7}\right]\right) .
\end{aligned}
$$

Using $\alpha-1n-\left(\frac{n}{3}+\frac{n}{5}+\frac{n}{7}\right)+\left(\frac{n}{3 \times 5}-1+\frac{n}{3 \times 7}-1+\frac{n}{5 \times 7}-1\right)-\frac{n}{3 \times 5 \times 7}, \\
10002178 \frac{3}{4},
\end{array} \text { so } 2178 \frac{3}{4}<n<2194 \frac{1}{16} .\right.
\end{array}
$$

Since $n$ is coprime with 105, the possible values of $n$ are $2179, 2182, 2183, 2186, 2188, 2189, 2192, 2194$. Upon verification, only $n=2186$ satisfies equation (1), hence the 1000th term of this sequence is 2186.
Solution 2 (Combinatorial Analysis Method) First, we need to understand the structure of the sequence in the problem, then proceed with the calculations.
First, we find the number of positive integers in $S=\{1,2,3, \cdots, 105\}$ that are coprime with $105=3 \times 5 \times 7$. Let $A_{i}=\{m \mid m \in S$ and $m$ is divisible by $i\} (i=3,5,7)$. Thus, the number of positive integers in $S$ that are coprime with 105 is
$$
\begin{aligned}
& \left|\bar{A}_{3} \cap \bar{A}_{5} \cap \bar{A}_{7}\right| \\
= & |S|-\left|A_{3}\right|-\left|A_{5}\right|-\left|A_{7}\right|+\left|A_{3} \cap A_{5}\right|+\left|A_{3} \cap A_{7}\right|+\left|A_{5} \cap A_{7}\right|-\left|A_{3} \cap A_{5} \cap A_{7}\right| \\
= & 105-\left[\frac{105}{3}\right]-\left[\frac{105}{5}\right]-\left[\frac{105}{7}\right]+\left[\frac{105}{3 \times 5}\right]+\left[\frac{105}{3 \times 7}\right]+\left[\frac{105}{5 \times 7}\right]-\left[\frac{105}{3 \times 5 \times 7}\right] \\
= & 105-35-21-15+7+5+3-1=48 .
\end{aligned}
$$

Next, let $\left\{a_{n}\right\}$ be the sequence of all positive integers that are coprime with 105, arranged in ascending order. Thus, $a_{1}=1, a_{2}=2, a_{3}=4, \cdots, a_{46}=101, a_{47}=103, a_{48}=104$, and let $P=\left\{a_{1}, a_{2}, \cdots, a_{48}\right\}$.
On one hand, each term $a_{n}$ in the sequence $\left\{a_{n}\right\}$ can be expressed as $a_{n}=105 \cdot k+r(k, r$ are non-negative integers, $0 \leqslant r \leqslant$ 104), thus, by $\left(a_{n}, 105\right)=(105 k+r, 105)=(r, 105)=1$, we know $r \in P$, i.e., each term in the sequence $\left\{a_{n}\right\}$ can be written in the form $a_{n}=105 k+r(k$ is a non-negative integer, $r \in P)$.

On the other hand, for any non-negative integer $k$ and any $r \in P$, by $(105 k+r, 105)=(r, 105)=1$, we know that there must be a term $a_{n}=105 k+r$ in the sequence $\left\{a_{n}\right\}$.

Thus, the sequence $\left\{a_{n}\right\}$ consists of and only of numbers of the form $105 k+r(k$ is a non-negative integer, $r \in P)$. For each fixed $k$, as $r$ takes all values in $P$, there are 48 numbers of the form $105 k+r$, forming 48 terms in the sequence. Since $1000=48 \times 20+40$, we have $a_{1000}=105 \times 20+a_{40}$.

And $a_{48}=104, a_{47}=103, a_{46}=101, a_{45}=97, a_{44}=94, a_{43}=92, a_{42}=89, a_{41}=88$, $a_{40}=86$, so $a_{1000}=105 \times 20+86=2186$.
Thus, the 1000th term of the sequence is 2186.