Three, (25 points) Given that the positive integer $M$ when divided by the positive integer $N$ leaves a remainder of 2, and the sum of the reciprocals of all values of $N$ is $\frac{627}{670}$, with the number of all values of $N$ being less than 16. Find all possible values of $M$.


Obviously, $N>2, N \mid (M-2)$.
Since $670=2 \times 5 \times 67$, $M-2$ has prime factors 2, 5, and 67. Therefore, the sum of the reciprocals of all positive divisors of $M-2$ is
$$
\frac{1}{1}+\frac{1}{2}+\frac{627}{670}=\frac{816}{335} .
$$
The number of positive divisors of $M-2$ is less than 18.
Since $(1+1)^{5}=32>18$, $M-2$ cannot have more than four prime factors.
If $M-2$ has only two prime factors, let
$$
\begin{array}{l}
M-2=2^{\alpha_{1}} \times 5^{\alpha_{2}} \times 67^{\alpha_{3}} . \\
\text { Then }\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right) < 18 . \\
\text { Hence }(1+2)(1+5)(1+67)(1+p) \\
=\frac{816}{335} \times 2 \times 5 \times 67 \times p .
\end{array}
$$

Solving for $p$ gives $p=3$.
Thus, $M-2=2 \times 5 \times 67 \times 3 \Rightarrow M=2010$.