Given that the greatest common divisor of $M$ and $N$ is $q^{q} r$, and it has $2(q+1)$ positive divisors, we have:
$$
\begin{array}{l}
q^{p}-1=2(q+1) \Rightarrow q\left(q^{p-1}-2\right)=3 \\
\Rightarrow q \mid 3 \Rightarrow q=3 \Rightarrow p=2 .
\end{array}
$$

Let $M=\overline{a b c d}$
$$
=1000 a+100 b+10 c+d=2 \times 3^{3} r \text {. }
$$

Then $N=\overline{d c b a}$
$$
=1000 d+100 c+10 b+a=3^{5} r \text {. }
$$

From equations (1) and (2), eliminating $r$ gives:
$$
818 a+80 b=10 c+181 d,
$$

or $10(82 a+8 b-c-18 d)=2 a+d$.
From equation (3), we know:
$$
\begin{array}{l}
818 a \leqslant 10 c+181 d \\
\leqslant 10 \times 9+181 \times 9=1719
\end{array}
$$
$\Rightarrow a<3 \Rightarrow a$ can be 1 or 2.
From equation (4), we know $10 \mid (2 a+d)$.
Thus, when $a=1$, $d=8$; when $a=2$, $d=6$.
(1) When $a=1, d=8$, from equation (4) we get:
$$
8 b=63+c \Rightarrow b=8+\frac{c-1}{8}
$$
$\Rightarrow c=1$ or 9, and correspondingly $b$ is 8 or 9.
Therefore, $M=1818$ or 1998.
From equation (1), we know $27 \mid M$. Thus, $M=1998$.
(2) When $a=2, d=6$, from equation (4) we get:
$$
c=8 b+55 \geqslant 55 \text {, }
$$

which contradicts $c<10$.
Therefore, the value of $M$ is 1998.
One. (20 points) Given two reverse-order four-digit numbers $M$ and $N$ with $q^{p}-1$ identical positive divisors, which can be factored into prime factors as $p q^{q} r$ and $q^{p+q} r$ respectively, where $p$, $q$, and $r$ are prime numbers. Find the value of $M$.
(Note: For example, 7284 and 4827 are reverse-order numbers. )