Three, Seeking Training: Find all natural numbers $n$ such that the numbers
$$
1,2, \cdots, 3 n
$$

can be arranged in a table
$$
\begin{array}{llll}
a_{1} & a_{2} & \cdots & a_{n} \\
b_{1} & b_{2} & \cdots & b_{n} \\
c_{1} & c_{2} & \cdots & c_{n}
\end{array}
$$

satisfying the following two conditions:
$$
\text { (1) } a_{1}+b_{1}+c_{1}=a_{2}+b_{2}+c_{2}=\cdots=a_{n}+b_{n}
$$
$+c_{n}$ and is a multiple of 6;
(2) $a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+\cdots+b_{n}=c_{1}$ $+c_{2}+\cdots+c_{n}$ and is a multiple of 6.
```
Let the set of natural numbers $n$ that satisfy the two conditions be denoted as $S$. Suppose $n \in S$, from conditions (1) and (2) we can deduce: there exist natural numbers $s$ and $t$, such that
$\qquad$
```
i.e., $3n+1=4s$,
$n(3n+1)=12t$.
Therefore, $n \equiv 1 \pmod{4}$,
$n \equiv 0 \pmod{3}$.
Thus, $n$ must be of the form
$12k+9, k=0,1,2,\cdots$
Proof 1: The smallest natural number $n$ that satisfies (1) is 9. We will prove that $9 \in S$. We have
It is easy to see that the 3 row sums and 3 column sums of $A_3$ are all 15. Furthermore, the 9 elements of $A_3$ are 1, 2, ..., 9. Now, let
$\alpha(3)=(1,8,6)$, $\beta(3)=(5,3,7)$, $\gamma(3)=(9,4,2)$, and construct the $3 \times 9$ number table $A$ as follows:
$A = \begin{array}{lllllllll} 1 & 8 & 6 & 23 & 21 & 25 & 18 & 13 & 11 \\ 14 & 12 & 16 & 9 & 4 & 2 & 19 & 26 & 24 \\ 27 & 22 & 20 & 10 & 17 & 15 & 5 & 3 & 7 \end{array}$
It is easy to see that the 27 elements of $A_9$ are exactly 1, 2, ..., 27, and the sum of each row is $3(15+9+18)=126 \equiv 0 \pmod{6}$.
Therefore, $9 \in S$, $S \neq \varnothing$.
Assume $m \in S$, we will prove that $9m \in S$.
Consider the table $A_m$, where each column sum is $6u$ and each row sum is $6v$, with $u$ and $v$ being natural numbers. Now, let the first row of $A_m$ be $\alpha(m)$, the second row be $\beta(m)$, and the third row be $\gamma(m)$, and construct the $3 \times 3m$ number table $A_{3m}$ as follows:
$A_{3m} = \begin{array}{lll} \alpha(m) & \beta(m) + 6m & \gamma(m) + 3m \\ \beta(m) + 3m & \gamma(m) & \alpha(m) + 6m \\ \gamma(m) + 6m & \alpha(m) + 3m & \beta(m) \end{array}$
where $\beta(m) + 3m$ denotes adding $3m$ to each element of $\beta(m)$, and the other notations have similar meanings. It is not difficult to see that the $9m$ elements of $A_{3m}$ are exactly 1, 2, ..., 9m, and
each column sum is $6u + 9m$,
each row sum is $18v + 9m^2$.
Now, let the first row, second row, and third row of $A_{3m}$ be denoted as $\alpha(3m)$, $\beta(3m)$, and $\gamma(3m)$, respectively, and construct the $3 \times 9m$ number table $A_{9m}$ as follows:
each column sum is $6u + 36m$, which is a multiple of 6.
each row sum is
$3(18v + 9m^2) + 3m \cdot 18m + 3m \cdot 9m = 54v + 108m$
This proves that if $m \in S$, then $9m \in S$.
In summary, $S \supset \{9^k | k=1,2,\cdots\}$, so $S$ is an infinite set.
Proof 2: We will prove
It is not difficult to see that the column sums of $A_m$ are all equal to $18k + 15$; the row sums of $A_m$ are all equal to $6k + 6$.
Next, we adjust $A_m$ to make the three row sums equal.
The first row of $A_m$ is an arithmetic sequence with a common difference $d_1 = 3$ and the first term $a_1 = 1$, so its $2l$-th term is
$a_{2l} = 1 + (2l-1) \cdot 3 = 6l - 2$.
The third row of $A_m$ is an arithmetic sequence with a common difference $d_3 = 3$ and the first term $c_2 = 6k + 3$, so its $2l$-th term is
$c_{2l} = 6k + 3 + (2l-1) \cdot 3 = 6l + 6k$.
It is easy to see that
$c_{2l} - a_{2l} = 1$.
Therefore, by swapping $a_{2l}$ and $c_{2l}$ in $A_m$, we can make the three row sums equal, all being
$(4k+3)(6k+5) = \frac{m(3m+1)}{2}$,
and the column sums remain unchanged, all being $18k + 15$.
Now, let the adjusted number table be denoted as $B_m$, and let the first row, second row, and third row of $B_m$ be denoted as $\alpha(m)$, $\beta(m)$, and $\gamma(m)$, respectively, and construct the $3 \times 9m$ number table $A_{9m}$ as follows:
$A_{9m} = \begin{array}{lll} \alpha(m) & \beta(m) + 6m & \gamma(m) + 3m \\ \beta(m) + 3m & \gamma(m) & \alpha(m) + 6m \\ \gamma(m) + 6m & \alpha(m) + 3m & \beta(m) \end{array}$
The column sums and row sums are all equal, and
column sum = $18k + 15 + 9m = 54k + 42 \equiv 0 \pmod{6}$,
row sum = $\frac{3}{2}m(3m+1) + 6m^2 + 3m^2 = 3(4k+3)(6k+5) + 9(4k+3) = 6(4k+3)(9k+7)$.
Assume $k \equiv 2 \pmod{9}$, and let $m = 4k + 3$. We first arrange 1, 2, ..., 9m such that
all are multiples of 6.
This shows that if $k \equiv 2 \pmod{9}$, then
$12k + 9 \in S$.
Therefore, $S \supset \{12k + 9 | k \equiv 2 \pmod{9}\}$ is an infinite set.
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