II. (This question is worth 25 points) Arrange all positive integers that are coprime with 105 in ascending order. Try to find the 1000th term of this sequence.
2186
II. 【Analysis and Solution】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 natural numbers that are coprime with 105 and no greater than 105 are $\left|\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3}\right|=|S|-\left(\mid A_{1}\right.$ $\left.|+| A_{2}|+| A_{3} \mid\right)+\left(\left|A_{1} \cap A_{2}\right|+\left|A_{1} \cap A_{2}\right|+\left|A_{2} \cap A_{3}\right|\right)-\left|A_{1} \cap A_{2} \cap A_{3}\right|$.
$\therefore|S|=105,\left|A_{1}\right|=35,\left|A_{2}\right|=21,\left|A_{3}\right|=15,\left|A_{1} \cap A_{2}\right|=7,\left|A_{1} \cap A_{4}\right|=5, \mid A_{1} \cap A_{2}$ $|=3,| A_{1} \cap A_{2} \cap A_{3} \mid=1$.
$\left|\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3}\right|=105-(3 \times 5+3 \times 7+5 \times 7)+(3+5+7)-1=105-71+15-1=48$.
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 1000
$$
\begin{aligned}
=48 & \times 20+40, \therefore a_{1000}=105 \times 220+a_{40} . \\
& \because a_{48}=104, a_{47}=103, a_{46}=101, a_{45}=97, a_{44}=94, a_{43}=92, a_{42}=89, a_{41}=88, a_{40}=86, \\
& \therefore a_{1000}=105 \times 20+86=2186 .
\end{aligned}
$$