Example 2 Arrange all positive integers that are coprime with 105 in ascending order, and find the 1000th term of this sequence.
(1994 National High School League Question)
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, $a_{1000}=2100+86=2186$.