II. Arrange all positive integers that are coprime with 105 in ascending order, and 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 $5 \mid x\}, A_{3}=\{x \mid x \in S$ and $7 \mid x\}$, then the number of natural numbers not greater than 105 and coprime with 105 is $\left|\overline{A_{1}} \cap \overline{A_{2}} \cap \overline{A_{3}}\right|=|S|-$
$$
\begin{array}{l}
\left(\left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|\right)+\left(\left|A_{1} \cap A_{2}\right|+\left|A_{1} \cap A_{2}\right|+\mid A_{2} \cap\right. \\
\left.A_{3} \mid\right)-\left|A_{1} \cap A_{2} \cap A_{3}\right| \text {. } \\
\therefore \quad|S|=105,\left|A_{1}\right|=35,\left|A_{2}\right|=21,\left|A_{3}\right|=15, \\
\left|A_{1} \cap A_{2}\right|=7,\left|A_{1} \cap A_{4}\right|=5, \quad\left|A_{2} \cap A_{3}\right|=3, \mid A_{1} \cap A_{2} \cap \\
A_{3} \mid=1 \text {. } \\
\left|\overline{A_{1}} \cap \overline{A_{2}} \cap \overline{A_{3}}\right|=105-(3 \times 5+3 \times 7+5 \times 7)+(3+ \\
5+7)-1 \\
=105-71+15-1=48 \text {. } \\
\end{array}
$$

Let the sequence of positive integers that are coprime with 105, arranged in ascending order, be $a_{1}, a_{2}, a_{3}, \cdots, a_{n} \cdots$.
Then $a_{1}=1, a_{2}=2, a_{3}=4, \cdots, a_{48}=104$.
Let $P=\left\{a_{1}, a_{2}, a_{3}, \cdots, a_{48}\right\}$. On one hand, for $n \geqslant 1$, let $a_{n}=$ $105 k+r(k \geqslant 0,0 \leqslant r \leqslant 104, k, r \in Z)$,
$\because\left(a_{n}, 105\right)=1, \therefore(r, 105)=1$, hence $r \in P$; on the other hand, for any non-negative integer $k$ and $r \in P, \because(r, 105)=1, \therefore(105 k+r$, 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
$$
\begin{array}{l}
1000=48 \times 20+40, \quad \therefore a_{1000}=105 \times 20+a_{40} \\
\because \quad a_{48}=104, \quad a_{47}=103, a_{46}=101, a_{45}=97, a_{44}=94 \text {, } \\
a_{43}=92, a_{42}=89, a_{41}=88, a_{40}=86 \text {, } \\
\therefore \quad a_{1000}=105 \times 20+86=2186 \text {. } \\
\end{array}
$$