Three. (25 points) Given a positive integer $M$ has $k$ positive divisors, among which, only two divisors are prime, and the sum of the reciprocals of these $k$ positive divisors is $\frac{342}{287}$. Find all values of $M$.

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\begin{array}{l}
\text{Because } 287=7 \times 41, \text{ so, 7 and 41 are exactly the two prime factors of } M. \\
\text{Let } M=7^{m} \times 41^{n}\left(m, n \in \mathbf{N}_{+}\right), d_{1}, d_{2}, \cdots, d_{k} \text{ be all the positive divisors of } M \text{ arranged in ascending order. Thus, } d_{1}=1, d_{k}=M. \text{ Then,} \\
\frac{342}{287}=\frac{1}{d_{1}}+\frac{1}{d_{2}}+\frac{1}{d_{3}}+\cdots+\frac{1}{d_{k}} \\
=\frac{d_{k}}{d_{k}}+\frac{d_{k-1}}{d_{k}}+\frac{d_{k-2}}{d_{k}}+\cdots+\frac{d_{1}}{d_{k}} \\
=\frac{d_{1}+d_{2}+d_{3}+\cdots+d_{k}}{d_{k}} \\
=\frac{\left(1+7+7^{2}+\cdots+7^{m}\right)\left(1+41+41^{2}+\cdots+41^{n}\right)}{7^{m} \times 41^{n}} . \\
\text{Hence, } \frac{7^{m+1}-1}{6} \cdot \frac{41^{n+1}-1}{40}=342 \times 7^{m-1} \times 41^{n-1} . \\
\text{By } \left(41^{n-1}, \frac{41^{n+1}-1}{40}\right)=1, \text{ we know that } \\
41^{n-1} \mid \frac{7^{m+1}-1}{6} . \\
\text{Let } 7^{m+1}-1=41^{n-1} \times 6 t\left(t \in \mathbf{N}_{+}\right) .
\end{array}
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Obviously, $(t, 7)=1$. Then,
$$
\frac{41^{n+1}-1}{40}=\frac{342}{t} \times 7^{m-1} \text{. }
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Therefore, $t \mid 342 \Rightarrow t \leqslant 342$.
Substituting equation (1) into equation (2) and rearranging, we get
$$
\begin{array}{l}
7^{m-1}\left(41^{2} \times 7^{2}-6 \times 13680\right)=41^{2}+6 t \\
\Rightarrow 289 \times 7^{m-1}=1681+6 t \\
\Rightarrow 1681<289 \times 7^{m-1} \leqslant 1681+6 \times 342 \\
\Rightarrow 5<\frac{1681}{289}<7^{m-1} \leqslant \frac{3733}{289}<13 \\
\Rightarrow m=2 .
\end{array}
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Substituting into equation (3), we get $t=57$.
Substituting into equation (1), we get $n=1$.
Thus, $M=7^{m} \times 41^{n}=2009$.
(Xie Wenxiao, Huanggang High School, Hubei, 438000)
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