$4 \cdot 24$ In a $3 \cdot 3$ square grid, fill in nine different natural numbers such that the product of the three numbers in each row and the product of the three numbers in each column are equal (we use $P$ to represent this product).
(1) Prove that this method of filling in the numbers is possible.
(2) Determine which of the six numbers $1990, 1991, 1992, 1993, 1994, 1995$ can be the value of $P$.
(3) Find the minimum value of $P$, and explain your reasoning.
(China Beijing Junior High School Grade 2 Mathematics Competition, 1991)
[Solution] (1) As shown in the figure, fill in the nine numbers, such that the product of the three numbers in each row and each column is 120.
(2) Clearly, the nine numbers filled in should be nine different divisors of $P$, and these nine divisors should not include $P$, so the number of divisors of $P$ is no less than 10.

Among the six numbers $1990, 1991, 1992, 1993, 1994, 1995$:
$1990=2 \cdot 5 \cdot 199$, which has 8 different divisors;
$1991=11 \cdot 181$, which has 4 different divisors;
$1992=2^{3} \cdot 3 \cdot 83$, which has 16 different divisors;
$1993=1 \cdot 1993$, which has 2 different divisors;
$1994=2 \cdot 997$, which has 4 different divisors;
$1995=3 \cdot 5 \cdot 7 \cdot 19$, which has 16 different divisors.
Clearly, $P$ cannot be $1990, 1991, 1993, 1994$.
$P$ could be 1992 or 1995, and
$$
P=1992 \text { when } \quad P=1995 \text { when }
$$
(3) In general, in the $2 \times 2$ square in the upper left corner of the grid, fill in four different natural numbers $a, b, c, d$. In the bottom right corner, fill in a natural number $e$ that is different from them.

The remaining four cells are filled with $\frac{c d}{e}, \frac{a b}{e}, \frac{b d}{e}, \frac{a c}{e}$, so that
$$
P=\frac{a b c d}{e} .
$$

Therefore, we only need to choose $a, b, c, d$ such that $\frac{c d}{e}, \frac{a b}{e}, \frac{b d}{e}, \frac{a c}{e}$ are all natural numbers, and they are all distinct.

In general, we can assume that $e$ is less than $a, b, c, d$. If not, we can appropriately swap rows and columns to move the smallest number to the bottom right corner, which does not affect the value of $P$.

Taking $e=1$, the product $P=\frac{a b c d}{e}$ takes the minimum value, which is clearly $2 \cdot 3 \cdot 4 \cdot 5=120$, as shown in the table in (1).
When $e=2$, $P \geqslant 180$.
When $e=3$, $P \geqslant 280$.
When $e=4$, $P \geqslant 420$.
When $e>5$, $\frac{a b c d}{e}>\frac{e^{4}}{e}>125>120$.
Therefore, the minimum value that $P$ can take is 120.