For a positive integer $n$, let $D(n)$ denote the set of positive integers not exceeding $n$ and coprime with $n$. For example, $D(12)=\{1,5,7,11\}$. Find all positive integers $n$ such that all elements in $D(n)$ can be arranged into an arithmetic sequence.
The set of positive integers $n$ sought is $\{n \mid n$ is a prime number not less than 5 or $\left.n=2^{t}\left(t \geqslant 3, t \in \mathbf{N}_{+}\right)\right\}$.
(1) Clearly, $n \leqslant 4$ does not meet the requirement, so $n>4$.
(2) If $n$ is an odd number greater than 4, then $1,2 \in D(n)$.
The elements in $D(n)$ can be arranged into an arithmetic sequence, so $1,2, \cdots, n-1 \in D(n)$.
Therefore, $D(n)$ meets the requirement if and only if $n$ is an odd prime.
(3) If $n=4 k(k \geqslant 2, k \in \mathbf{N})$, then $2 k-1, 2 k+1 \in D(n)$.
Since $(4 k-1, n)=1$, it follows that
$4 k-1 \in D(n)$, and $1 \in D(n)$.
If all elements in $D(n)$ can be arranged into an arithmetic sequence, it indicates that $D(n)$ consists of all odd numbers less than $n$. Thus, $k$ cannot have any odd prime factors, and $k$ can only be a power of 2.

It is easy to verify that such $n=2^{t}\left(t \geqslant 3, t \in \mathbf{N}_{+}\right)$ all meet the requirement.
(4) If $n=2 m$ and $m \geqslant 5$ is an odd number, then $m-2, m-4 \in D(n)$.
Since $(2 m-1, n)=1$, it follows that
$2 m-1 \in D(n)$, and $1 \in D(n)$.
If all elements in $D(n)$ can be arranged into an arithmetic sequence, it indicates that $D(n)$ consists of all odd numbers less than $n$. Thus, $m \in D(n)$, but $m \notin D(n)$, so $n$ does not meet the requirement.
(5) Also, $D(6)=\{1,5\}$, so $n=6$ does not meet the requirement.

In summary, the set of positive integers $n$ sought is $\{n \mid n$ is a prime number not less than 5 or $\left.n=2^{t}\left(t \geqslant 3, t \in \mathbf{N}_{+}\right)\right\}$.