For positive integers $m, n$, if every row of the $m \times n$ table is a permutation of $1,2, \cdots, n$, and each column, after appropriate rearrangement, is an arithmetic sequence of length $m$ with a common difference greater than 0, then the table is called a continuous arithmetic table of size $m \times n$.
Find all positive integers $n$ such that there exists a $3 \times n$ continuous arithmetic table.
Clearly, $n \geqslant 3$, because in a $3 \times n$ continuous arithmetic number table, the 3 numbers in each column are distinct, so the number table contains at least 3 different numbers.
$n=3$ satisfies the requirement. At this time,
$$
\left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1 \\
3 & 1 & 2
\end{array}\right)
$$

is a $3 \times 3$ continuous arithmetic number table that meets the requirement.
Moreover, if $n=k$ satisfies the requirement, then $n=k+3$ also satisfies the requirement.
In fact, when $n=k$, the $3 \times k$ continuous arithmetic number table is
$$
\left(\begin{array}{cccc}
a_{1} & a_{2} & \cdots & a_{k} \\
b_{1} & b_{2} & \cdots & b_{k} \\
c_{1} & c_{2} & \cdots & c_{k}
\end{array}\right)
$$

then
$$
\left(\begin{array}{lllllll}
a_{1} & a_{2} & \cdots & a_{k} & k+1 & k+2 & k+3 \\
b_{1} & b_{2} & \cdots & b_{k} & k+2 & k+3 & k+1 \\
c_{1} & c_{2} & \cdots & c_{k} & k+3 & k+1 & k+2
\end{array}\right)
$$

is a $3 \times (k+3)$ continuous arithmetic number table.
When $n=4$, if there exists a $3 \times 4$ continuous arithmetic number table, then each column is a length-3 arithmetic sequence composed of 3 numbers from $1,2,3,4$. Such a sequence must contain 2, so 2 appears in each column, appearing 4 times in total. But the number table has only 3 rows, and each number appears only once in the same row, appearing 3 times in total, which is a contradiction. Therefore, $n=4$ does not satisfy the requirement.

When $n=5$, if there exists a $3 \times 5$ continuous arithmetic number table, then each column is a length-3 arithmetic sequence composed of 3 numbers from $1,2,3,4,5$. Since 1 appears in each row, appearing 3 times in total, there are 3 arithmetic sequences containing 1, and such sequences must contain 3, so 3 appears in 3 arithmetic sequences containing 1. Similarly, there are 3 arithmetic sequences containing 4, and such sequences must contain 3, so 3 appears in 3 arithmetic sequences containing 4. But one such arithmetic sequence cannot contain both 1 and 4, so 3 appears at least $6>3$ times, which is a contradiction.
When $n=6$, since $n=3$ satisfies the requirement, $n=3+3=6$ also satisfies the requirement.
When $n=7$, there exists a $3 \times 7$ continuous arithmetic number table
$$
\left(\begin{array}{lllllll}
1 & 2 & 3 & 4 & 5 & 6 & 7 \\
3 & 4 & 2 & 6 & 7 & 5 & 1 \\
5 & 6 & 1 & 2 & 3 & 7 & 4
\end{array}\right)
$$

When $n=8$, there exists a $3 \times 8$ continuous arithmetic number table
$$
\left(\begin{array}{llllllll}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
2 & 5 & 1 & 7 & 6 & 8 & 3 & 4 \\
3 & 8 & 2 & 1 & 4 & 7 & 5 & 6
\end{array}\right)
$$

By mathematical induction, the set of all positive integers $n$ for which there exists a $3 \times n$ continuous arithmetic number table is $\{n \in \mathbf{Z} \mid n=3 \text{ or } n \geqslant 6\}$.