Example 2 Arrange all positive integers that are coprime with 105 in ascending order, and find the 1000th term of this sequence.
Let the sequence be denoted as $a_{1}, a_{2}, a_{3}, \cdots$. The number of positive integers not exceeding 105 and coprime with 105 is $105-\left(\frac{105}{3}+\frac{105}{5}+\frac{105}{7}\right)+\left(\frac{105}{15}+\frac{105}{35}+\frac{105}{21}\right)-1=48$.
Since $1000=20 \cdot 48+40$, it follows that $a_{1000}=20 \cdot 105+a_{40}=2100+a_{40}$.
Furthermore, $a_{48}=104, a_{47}=103, a_{46}=101, a_{45}=97$,
$a_{44}=94, a_{43}=92, a_{42}=89, a_{41}=88, a_{40}=86$.
Therefore, $a_{1000}=2100+86=2186$.