Three. (25 points) If placing the natural number $N$ to the right of any natural number always results in a new number that is divisible by $N$, then $N$ is called a "magic number." Find all magic numbers.

---

Translation:

Three. (25 points) If placing the natural number $N$ to the right of any natural number always results in a new number that is divisible by $N$, then $N$ is called a "magic number." Find all magic numbers.
Let $N$ be a magic number, with the number of digits being $m$. Then, the new number obtained by placing $N$ to the right of 1 is $10^{m}+N$.

Since $N \perp\left(10^{m}+N\right)$, it follows that $N \perp 10^{m}$. Therefore, $N=2^{a} \cdot 5^{b}$ (where $a$ and $b$ are non-negative integers not exceeding $m$), which means
$$
N=2^{r} \cdot 10^{k} \text { or } 10^{k} \text { or } 5^{r} \cdot 10^{k}
$$
($k$ is the smaller of $a$ and $b$, $r$ is a positive integer, $r+k \leqslant m$ and $r+k$ is the larger of $a$ and $b$).

When $N=5^{r} \cdot 10^{k}$, since $5^{4}=625$, if $r \geqslant 4$, then $5^{r}=5^{4} \cdot 5^{r-4}$, and its number of digits $\leqslant 3+$ $(r-4)=r-1$. Therefore, the number of digits of $5^{r} \cdot 10^{k}$ $\leqslant r-1+$ $k<r+k \leqslant m$, which is a contradiction.
Hence, $r \leqslant 3$, i.e., $N=5 \cdot 10^{k}, 5^{2} \cdot 10^{k}, 5^{3} \cdot 10^{k}$.
Similarly, when $N=2^{r} \cdot 10^{k}$, only $N=2 \cdot 10^{k}$ is a magic number.

Conversely, $N$ is a magic number if it is one of $2 \cdot 10^{k}, 5^{r} \cdot 10^{k}(r=0,1,2,3, k$ being a non-negative integer). In this case, $N$ is the desired number.