II. (This question is worth 25 points) Arrange all positive integers that are coprime with 105 in ascending order. Find the 1000th term of this sequence.
$$
\begin{array}{r}
\text { 2. Let } S=\{1,2, \cdots, 105\}, A_{1}=\{x \mid x \in S, \text { and } 3 \mid \\
x\}, A_{2}=\{x \mid x \in S, 45 \mid x\}, A_{3}=\{x \mid x \in S, \text { and } 7 \mid
\end{array}
$$
$$
\begin{array}{l}
\left|\overline{A_{1}} \cap \overline{A_{2}} \cap \overline{A_{3}}\right|=| S \mid-\left(\left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|\right) \\
+\left(\left|A_{1} \cap A_{2}\right|+\left|A_{1} \cap A_{3}\right|+\left|A_{2} \cap A_{3}\right|\right) \\
-\left|A_{1} \cap A_{2} \cap A_{3}\right| \text {. } \\
\because|S|=105,\left|A_{1}\right|=35,\left|A_{2}\right|=21,\left|A_{3}\right|=15, \\
\left|A_{1} \cap A_{2}\right|=7,\left|A_{1} \cap A_{3}\right|=5,\left|A_{2} \cap A_{3}\right|=3, \\
\left|A_{1} \cap A_{2} \cap A_{3}\right|=1, \\
\therefore\left|\overline{A_{1}} \cap \overline{A_{2}} \cap \overline{A_{3}}\right|=105-(3 \times 5+3 \times 7+5 \times 7) \\
+(3+5+7)-1 \\
=105-71+15-1=48. \\
\end{array}
$$

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 $a_{n}=105 k+r(k \geqslant 0,0 \leqslant r \leqslant 104, k, r \in \mathbb{Z})$. Since $(a_{n}, 105)=1$, we have $(r, 105)=1$, hence $r \in P$; on the other hand, for any non-negative integer $k$ and $r \in P$, since $(r, 105)=1$, we have $(105 k+r, 105)=1$. Therefore, there must be an $n$ such that $a_{n}=105 k+r$. This shows that the sequence $\left\{a_{n}\right\}$ consists of and only of numbers of the form $105 k+r(k \geqslant 0, k \in \mathbb{Z}, r \in P)$ arranged in ascending order.

Since the sequence is strictly increasing, and for each fixed non-negative integer $k$, when $r$ takes the values in $P$, there are 48 numbers, and
$$
\begin{array}{c}
1000=48 \times 20+40, \text { so } a_{1000}=105 \times 20+a_{40}. \\
\because a_{48}=104, a_{47}=103, a_{46}=101, a_{45}=97, \\
a_{44}=94, a_{43}=92, a_{42}=89, a_{41}=88, a_{40}=86, \\
\therefore a_{1000}=105 \times 20+86=2186.
\end{array}
$$