Example 2 If a natural number $N$ is appended to the right of any natural number, the resulting number can be divided by $N$ (for example, 2 appended to 35 results in 352, which is divisible by 2), then $N$ is called a "magic number". Among the natural numbers less than 130, how many magic numbers are there?
Analysis: To calculate how many magic numbers there are, we first need to clarify what constitutes a magic number. Although the problem provides the definition of a magic number, it is not convenient to directly use this definition to determine whether a number is a magic number. Therefore, we need to explore a simplified and equivalent condition.

Obviously, the key to simplification is how to accurately describe in mathematical language "a number appended to the right of a natural number." Imagine $P$ appended to the right of $Q$; what is the resulting number? How can this number be expressed in terms of $P$ and $Q$? This clearly depends on how many digits $P$ has. If $P$ is an $m$-digit number, then the number obtained by appending $P$ to the right of $Q$ is $Q \cdot 10^{m} + P$. Using this, we can derive a simplified and equivalent condition based on the definition of a magic number.

Solution: Let $P$ be a magic number, and assume it is an $m$-digit number.
According to the condition, for any natural number $Q$, we have
$$
P \mid\left(Q \cdot 10^{m} + P\right),
$$

which implies $P \mid Q \cdot 10^{m}$ (for any $Q \in \mathbf{N}$).
Taking $Q=1$, we get $P \mid 10^{m}$.
Conversely, if $P \mid 10^{m}$, then for any $Q \in \mathbf{N}$, we have $P \mid Q \cdot 10^{m}$.
Thus, $P \mid\left(Q \cdot 10^{m} + P\right)$.
Therefore, $P$ is an $m$-digit magic number if and only if $P \mid 10^{m}$.
(1) When $m=1$, $P \mid 10$, and $P$ is a one-digit number, so $P=1,2,5$.
(2) When $m=2$, $P \mid 10^{2}$, i.e., $P \mid\left(2^{2} \times 5^{2}\right)$, and $P$ is a two-digit number, so,
$$
P=2^{0} \times 5^{2}, 2^{1} \times 5^{1}, 2^{1} \times 5^{2}, 2^{2} \times 5^{1}.
$$
(3) When $m=3$, $P \mid 10^{3}$, i.e., $P \mid\left(2^{3} \times 5^{3}\right)$, and $P$ is a three-digit number, $P<130$, so,
$$
P=2^{0} \times 5^{3}, 2^{2} \times 5^{2}.
$$

In summary, there are $3+4+2=9$ magic numbers that meet the conditions.