Example 7 Find all positive integers $n$ that satisfy the following condition: there exist two complete residue systems modulo $n$, $a_{i}$ and $b_{i} (1 \leqslant i \leqslant n)$, such that $a_{i} b_{i} (1 \leqslant i \leqslant n)$ is also a complete residue system modulo $n$.
[2]
【Analysis and Solution】When $n=1,2$, let $a_{i}=b_{i}=i$ $(1 \leqslant i \leqslant n)$, so $n=1,2$ satisfy the condition.
The following proves that $n \geqslant 3$ does not satisfy the condition.
(1) An odd prime $p$ does not satisfy the condition.
In fact, suppose $a_{i} 、 b_{i} 、 a_{i} b_{i}(1 \leqslant i \leqslant p)$ respectively traverse the complete system modulo $p$. Without loss of generality, assume $p \mid a_{1} b_{1}$, then when $2 \leqslant i \leqslant p$, $p \nmid a_{i} b_{i}$.
Thus, $p \nmid a_{i}, p \nmid b_{i}$.
Therefore, $a_{i} 、 b_{i} 、 a_{i} b_{i}(2 \leqslant i \leqslant p)$ respectively traverse the reduced system modulo $p$.
By Wilson's theorem and property 4, we have
$$
\begin{array}{l}
(-1)^{2} \equiv[(p-1)!]^{2} \equiv\left(\prod_{i=2}^{p} a_{i}\right)\left(\prod_{i=2}^{p} b_{i}\right) \\
=\prod_{i=2}^{n}\left(a_{i} b_{i}\right) \equiv-1(\bmod p),
\end{array}
$$

which implies $p \nmid 2$, a contradiction.
(2) Transfer the method in (1) to a general odd number $n(n \geqslant 3)$.
Let $p$ be a prime factor of the odd number $n$, $n=p m$.
Still assume $a_{i} 、 b_{i} 、 a_{i} b_{i}$ respectively traverse the complete system modulo $p m$.
Without loss of generality, assume $a_{1} b_{1}, a_{2} b_{2}, \cdots, a_{m} b_{m}$ are all the multiples of $p$ in $a_{i} b_{i}(1 \leqslant i \leqslant p m)$. Then when $m+1 \leqslant i \leqslant p m$, $p \nmid a_{i} b_{i}$.
Thus, $p \nmid a_{i}, p \nmid b_{i}$.
Therefore, $a_{i} 、 b_{i} 、 a_{i} b_{i}(m+1 \leqslant i \leqslant p m)$ must respectively traverse $p j+i(1 \leqslant i \leqslant p-1,0 \leqslant j \leqslant m-1)$ modulo $p m$.
When $m$ is odd, since
$$
\begin{array}{l}
\prod_{i=1}^{p-1} \prod_{j=0}^{m-1}(p j+i) \equiv \prod_{i=1}^{p-1} i^{m}=[(p-1)!]^{m} \\
\equiv(-1)^{m}=-1(\bmod p), \\
\text { hence } \prod_{i=m+1}^{m} a_{i} \equiv \prod_{i=m+1}^{m} b_{i} \equiv \prod_{i=m+1}^{m} a_{i} b_{i}
\end{array}
$$
$\equiv-1(\bmod p)$.
But the above is a contradiction.
Therefore, odd numbers $n(n \geqslant 3)$ do not satisfy the condition.
(3) From another perspective, for $n$ having a square prime factor $p^{2}$, a negative conclusion can also be drawn.
In fact, let $\frac{n}{p}=m$.
Without loss of generality, assume $a_{1} b_{1}, a_{2} b_{2}, \cdots, a_{m} b_{m}$ are all the multiples of $p$ in $a_{i} b_{i}(1 \leqslant i \leqslant n)$, then $p \nmid a_{i} b_{i}(m<i \leqslant n)$.
Thus, $p \nmid a_{i}, p \nmid b_{i}$.
Therefore, there must be $p\left|a_{i}, p\right| b_{i}(1 \leqslant i \leqslant m)$.
Thus, $a_{i} b_{i}(1 \leqslant i \leqslant m)$ are all multiples of $p^{2}$.
Therefore, there does not exist $i$ such that $a_{i} b_{i} \equiv p(\bmod n)$.
Hence, when $n$ has a square prime factor, the conclusion is negative.
(4) Now, only the case $n=2 p m$ ( $p$ is an odd prime, $(m, 2 p)=1, m$ has no square factor) needs to be studied.

In fact, by slightly modifying the method in (2), a negative answer can be given for the more general case of " $n=2 p m, p$ is an odd prime, $m$ is odd".

In fact, consider the set of all numbers from 1 to $n=2 p m$ that are coprime with $2 p$.
Notice,
$$
\begin{array}{l}
\prod_{\substack{1 \leq i \leq 1<p \\
(i, 2 p)=1}} i=1 \times 3 \times \cdots \times(p-2) \times \\
(p+2) \times(p+4) \times \cdots \times(2 p-1) \\
=1 \times 3 \times \cdots \times(p-2) \times 2 \times 4 \times \cdots \times(p-1) \\
=(p-1)!\equiv-1(\bmod p) \text {. } \\
\text { Hence } \prod_{\substack{1 \leq i \leq 2 p m \\
(i, 2 p)=1}} i=\prod_{\substack{1 \leq i \leq 2 p \\
(i, 2 p)=1}} \prod_{j=0}^{m-1}(2 p j+i) \\
\equiv \prod_{\substack{\leq i \leq i \leq 2 p \\
(i, 2 p)=1}} i^{m} \equiv(-1)^{m}=-1(\bmod p) \text {. } \\
\end{array}
$$

Now assume $a_{i} 、 b_{i} 、 a_{i} b_{i}$ respectively traverse the complete system modulo $2 p m$.

Without loss of generality, assume $a_{i} b_{i}(1 \leqslant i \leqslant t)$ are all the numbers coprime with $2 p$ in $a_{i} b_{i}(1 \leqslant i \leqslant 2 p m)$.
Then when $1 \leqslant i \leqslant t$, we have
$\left(2 p, a_{i}\right)=\left(2 p, b_{i}\right)=1$.
Therefore, $a_{i} 、 b_{i} 、 a_{i} b_{i}(1 \leqslant i \leqslant t)$ must respectively traverse $i(1 \leqslant i \leqslant 2 p m,(2 p, i)=1)$ modulo $2 p m$.
Notice equation (1), we know
$\prod_{i=1}^{t} a_{i} \equiv \prod_{i=1}^{t} b_{i} \equiv \prod_{i=1}^{i} a_{i} b_{i} \equiv-1(\bmod p)$.
But the above is a contradiction.
Combining (2) and (4), we know that all integers $n \geqslant 3$ do not satisfy the condition.
Therefore, only $n=1,2$ satisfy the condition.