Let the set of natural numbers $n$ that satisfy the above two conditions be denoted as $s$. If $n \in s$, by conditions (1) and (2), we know: there exist natural numbers $s$ and $t$, such that
$$\left\{\begin{array}{l} \frac{3 n(3 n+1)}{2}=6 s n \\ \frac{3 n(3 n+1)}{2}=18 t \end{array}\right.$$

which implies $\left\{\begin{array}{l}3 n+1=4 s \\ n(3 n+1)=12 t\end{array}\right.$
Thus, $\left\{\begin{array}{ll}n \equiv 1 & (\bmod 4) \\ n \equiv 0 & (\bmod 3)\end{array}\right.$
Therefore,
$n=12 k+9, k$ is a non-negative integer.
Conversely, if $n=12 k+9, k=0,1,2, \cdots$, and the sum of the rows and the sum of the columns in the number table formed by $1,2, \cdots, 3 n$ are all equal, then
$$\begin{array}{l} \text { Column sum }=\frac{3(3 n+1)}{2}=\frac{3(36 k+28)}{2}=6(9 k+7), \\ \text { Row sum }=\frac{n(3 n+1)}{2}=\frac{n}{3} \cdot \frac{3(3 n+1)}{2}=(4 k+3) \cdot 6 \cdot(9 k+7) . \end{array}$$

It is evident that both the column sum and the row sum are multiples of 6.
Next, we will prove that when $n=9^{k}, k=1,2, \cdots$, the numbers $1,2, \cdots, 3 n$ can be arranged into a $3 \times n$ number table such that the row sums are all equal and the column sums are all equal.

When $k=1$, $n=9, 3 n=27$.
At this point, note that
$$\left(\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{array}\right)+\left(\begin{array}{lll} 0 & 6 & 3 \\ 3 & 0 & 6 \\ 6 & 3 & 0 \end{array}\right)=\left(\begin{array}{lll} 1 & 8 & 6 \\ 5 & 3 & 7 \\ 9 & 4 & 2 \end{array}\right)$$

The $3 \times 3$ number table on the right side of the above equation has row sums all equal to 15, and column sums all equal to 15. Denote the three rows of this $3 \times 3$ number table as
$$\alpha(3), \beta(3), \gamma(3)$$

i.e., $\alpha(3)=(1,8,6), \beta(3)=(5,3,7), \gamma(3)=(9,4,2)$. Construct a $3 \times 9$ number table
$$A_{9}=\left(\begin{array}{ccc} \alpha(3) & \beta(3)+18 & \gamma(3)+9 \\ \beta(3)+9 & \gamma(3) & \alpha(3)+18 \\ \gamma(3)+18 & \alpha(3)+9 & \beta(3) \end{array}\right)$$

i.e.,
$$A_{9}=\left(\begin{array}{ccccccccc} 1 & 8 & 6 & 23 & 21 & 25 & 18 & 13 & 11 \\ 14 & 12 & 16 & 9 & 4 & 2 & 19 & 26 & 24 \\ 27 & 22 & 20 & 10 & 17 & 15 & 5 & 3 & 7 \end{array}\right)$$

The row sums of this number table are clearly equal, and the column sums are also clearly equal.
Assume that when $m=9^{k}$, the numbers $1,2, \cdots, 3 m$ can be arranged into a $3 \times m$ number table
$$A_{m}=\left(\begin{array}{llll} a_{1} & a_{2} & \cdots & a_{m} \\ b_{1} & b_{2} & \cdots & b_{m} \\ c_{1} & c_{2} & \cdots & c_{m} \end{array}\right)$$

such that the row sums of $A_{m}$ are all equal, and the column sums are all equal.
Then, when $n=9 m=9^{k+1}$, we can construct a $3 \times 3 m$ number table as follows:
$$A_{3 m}=\left(\begin{array}{ccc} \alpha(m) & \beta(m)+6 m & \gamma(m)+3 m \\ \beta(m)+3 m & \gamma(m) & \alpha(m)+6 m \\ \gamma(m)+6 m & \alpha(m)+3 m & \beta(m) \end{array}\right)$$

where $\alpha(m), \beta(m), \gamma(m)$ are the first, second, and third rows of $A_{m}$, respectively. Clearly, the $9 m$ elements in $A_{3 m}$ are exactly $1,2, \cdots, 9 m$, and the column sums of $A_{3 m}$ are all equal, and the row sums of $A_{3 m}$ are all equal.

Next, we construct a $3 \times n$ number table, i.e., a $3 \times 9 m$ number table as follows:
$$A_{n}=A_{9 m}=\left(\begin{array}{ccc} \alpha(3 m) & \beta(3 m)+18 m & \gamma(3 m)+9 m \\ \beta(3 m)+9 m & \gamma(3 m) & \alpha(3 m)+18 m \\ \gamma(3 m)+18 m & \alpha(3 m)+9 m & \beta(3 m) \end{array}\right)$$

where $\alpha(3 m), \beta(3 m), \gamma(3 m)$ are the first, second, and third rows of $A_{3 m}$, respectively. Clearly, the column sums of $A_{9 m}$ are all equal, and the row sums of $A_{9 m}$ are all equal, and the $27 m$ elements in $A_{9 m}$ are exactly $1,2, \cdots, 27 m$.

According to the principle of mathematical induction, when $n=9^{k}, k=1,2, \cdots$, the numbers $1,2, \cdots, 3 n$ can be arranged into a $3 \times n$ number table such that the row sums are all equal and the column sums are all equal.

Notice that when $k=1$, $9^{k}=9$; and when $k>1$,
$$\begin{aligned} n & =9^{k}=9^{k}-9+9=9 \cdot\left(9^{k-1}-1\right)+9 \\ & =9 \cdot 8 \cdot m+9 \\ & =12 \cdot 6 m+9 \end{aligned}$$

where $m$ is an integer. Therefore, when $n=9^{k}, k=1,2, \cdots$, $n$ can always be expressed in the form $n=12 l+9$ (where $l$ is an integer). Thus, the row sums and column sums of the number table $A_{n}$ are all multiples of 6.

Hence, the original proposition is proved.