Two, (50 points) The four-digit numbers $m$ and $n$ are reverse positive integers of each other, and $m+n=18 k+9\left(k \in \mathbf{N}_{+}\right), m$ and $n$ have 16 and 12 positive divisors (including 1 and themselves), respectively. The prime factors of $n$ are also prime factors of $m$, but $n$ has one fewer prime factor than $m$. Find all possible values of $m$.
Let $m=\overline{a b c d}, a d \neq 0$. Then $n=\overline{d c b a}$.
Given $m+n=9(2 k+1)$, then $9 \mid (m+n)$.
Thus, $9 \mid [(1000 a+100 b+10 c+d)+$
$$
\begin{array}{c}
(1000 d+100 c+10 b+a)], \\
9 \mid 2(a+b+c+d), 9 \mid (a+b+c+d).
\end{array}
$$

Therefore, $9 \mid m, 9 \mid n$.
Since $m+n$ is odd, it follows that $m$ and $n$ are of opposite parity.
If $n$ is even, i.e., $2 \mid n$, then $2 \mid m$, making $m$ even, which is a contradiction.
Thus, $m$ is even, and $n$ is odd.
Let the number of 3's in the prime factorization of $m$ be $\alpha_{1}$, and the number of 2's be $\alpha_{2}$. Then $\alpha_{1} \geqslant 2, \alpha_{2} \geqslant 1$.
By the divisor count theorem,
$\left[\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\right] \mid 16$.
Thus, $\left(\alpha_{1}+1\right) \mid 8, \alpha_{1}+1 \geqslant 3$.
So, $\alpha_{1}+1=4$ or $8, \alpha_{1}=3$ or 7.
Hence, $m$ has at most three prime factors.
Thus, $n$ has at most two prime factors, and 3 is one of the prime factors of $n$.

If $n$ has only one prime factor, then this prime factor is 3. Thus, $n=3^{11}>10000$, which contradicts the fact that $n$ is a four-digit number.
Therefore, $n$ has two prime factors.
Let the other prime factor of $n$ be $p$.
Since $9 \mid n$, we have
$n=3^{2} p^{3}$ or $3^{3} p^{2}$ or $3^{5} p(p>3)$.
Thus, $m=2^{\alpha_{2}} 3^{\alpha_{1}} p^{\alpha_{3}}\left(\alpha_{1}, \alpha_{2}, \alpha_{3} \in \mathbf{N}_{+}\right)$.
Also, $\left(\alpha_{2}+1\right)\left(\alpha_{1}+1\right)\left(\alpha_{3}+1\right)=16$, so $\alpha_{1}=3, \alpha_{2}=\alpha_{3}=1$, i.e., $m=54 p$.
Since $m \geqslant 1000$, we have $p \geqslant 19$.
At this point, $3^{2} p^{3}$ is greater than 9999.
When $p=19$, $3^{3} p^{2}=9747$.
But $m=54 p=1026$ are not reverse numbers, so $p \geqslant 23$. At this point, $3^{3} p^{2}>9999$.
Therefore, $n=3^{5} p$. Thus,
$\frac{m}{n}=\frac{2}{9} \Rightarrow 9 m=2 n$,
$9(1000 a+100 b+10 c+d)$
$=2(1000 d+100 c+10 b+a)$,
$818 a+80 b=181 d+10 c$.
$818 a \leqslant 181 \times 9+10 \times 9=1719<3 \times 818$.
Thus, $a<3$.
Since $n$ is odd, $a$ must be odd. Hence, $a=1$.
From equation (1),
$d=\frac{818+80 b-10 c}{181} \geqslant \frac{818-10 \times 9}{181}=4 \frac{4}{181}$.
Since $m$ is even, $d$ must be even.
Thus, $d=6$ or 8.
When $d=6$, from equation (1),
$880 b-110 c=2948$.
Since $51 \mid (80 b-10 c)$, we have $d=8$.
Thus, $8 b-c=63, 8 b=63+c \geqslant 63, b \geqslant 7 \frac{7}{8}$.
Thus, $b=8$ or 9.
When $b=8$, $c=1$;
When $b=9$, $c=9$.
Thus, $m=1818$ or 1998.
Since $271 \mid m$, $m \neq 1818$.
Also, $m=1998=2 \times 3^{3} \times 37, n=8991=$ $3^{5} \times 37$ satisfies the conditions.
Therefore, $m=1998$.