$8 \cdot 5$ List all positive integers that are coprime with 105 in ascending order, and find the 1000th term of this sequence.
[Solution] 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 $1000=20 \times 48+40$, we have
$$
a_{1000}=20 \times 105+a_{40}=2100+a_{40} .
$$

Furthermore, given that $a_{48}=104, a_{47}=103, a_{46}=101, a_{45}=97, a_{44}=94, a_{43}=92$,
$$
a_{42}=89, a_{41}=88, a_{40}=86 \text {, }
$$

it follows that $a_{1000}=2186$.