Three, in a $3 \times 3$ 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 all equal (we denote this product by $P$).

(1) Prove that this arrangement of 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.
Three, (1) Proof It is easy to prove the following filling method 1 As shown in the figure, fill in 1, 2, 3, 4, 5, 6, $8, 15, 20$ nine different natural numbers, then the product of the three numbers in each row and the product of the three numbers in each column are all equal to 120. 
Therefore, the filling method required by the problem can be achieved.
(2) Solution Obviously, the nine natural numbers to be filled should be nine different divisors of $P$.
Also, the divisor $P$ itself cannot be filled into the table,
Therefore, if the filling method can be achieved, the number of different divisors of $P$ must be greater than or equal to 10.

Among 1990, 1991, 1992, 1993, 1994, 1995, $1990=2 \times 5 \times 199$ has a total of eight different divisors. $1991=11 \times 181$ has a total of four different divisors, $1992=2^{8} \times 3 \times 83$ has a total of sixteen different divisors, $1993=1 \times 1993$ has a total of two different divisors, $1994=2 \times 997$ has a total of four different divisors, $1995=3 \times 5 \times 7 \times 19$ has a total of sixteen different divisors.

Obviously, $P$ cannot take 1990, 1991, 1993, 1994, $P$ may take the values 1992, 1995.
Also, for $P=1992$, there is 1.
$$
P=1995 \text { has } 2 \text {. }
$$

Therefore, among 1990, 1991, 1992, 1993, 1994, 1995, the possible values of $P$ are 1992 and 1995. (3) Solution In general, in the $2 \times 2$ square in the upper left corner, fill in four different natural numbers $a$, $b$, $c$, $d$. In the lower right corner, fill in a natural number $e$ that is different from them, and in the remaining four cells, fill in $-\frac{d}{e}$, $-\frac{a b}{e}$, $\frac{b d}{e}$, $\frac{a c}{e}$, so that $P=\frac{a b c d}{e}$. In this case, we only need to select $a$, $b$, $c$, $d$ 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 different from each other.

In general, it can be assumed that $e$ is smaller than $a$, $b$, $c$, $d$. If not, appropriately swap rows and columns to move the smallest number to the lower right corner, without affecting the value of $P$. When $e=1$, the product $P=\frac{a b c d}{e}$ takes the minimum value, which is clearly 120, as shown in the table filled in (1).
When $e=2$, $P \geqslant 180$,
When $e=3$, $P \geqslant 280$,
When $e=4$, $P \geqslant 420$.
Finally, 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.