12. Let $M$ be a subset of the set $N=\{1,2,3, \cdots, 1998 \mid$ such that each natural number (element) in $M$ has exactly 1 zero. Then the set $M$ can contain at most $\qquad$ elements.
12.414

Divide the set $S$ into 4 subsets:
$$
\begin{array}{l}
S_{1}=\{1,2, \cdots, 9\} \\
S_{2}=\{10,11, \cdots, 99\} \\
S_{3}=\{100,101, \cdots, 999\} \\
S_{4}=\{1000,1001, \cdots, 1998\}
\end{array}
$$

Let $M_{i}$ denote the subset of $S_{i}(I=1,2,3,4)$ consisting of all elements that contain exactly one digit 0. Clearly, none of the elements in $S_{1}$ contain a 0, so $\left|M_{1}\right|=0$;
In $S_{2}$, there are 9 elements with 0 in the units place, so $\left|M_{2}\right|=9$;
In $S_{3}$, there are $9^{2}$ elements with 0 in the units or tens place, so $\left|M_{3}\right|=2 \times 9^{2}=162$.
In $S_{4}$, there are $9^{2}$ elements with 0 in the units, tens, or hundreds place, so $\left|M_{4}\right|=3 \times 9^{2}=243$.
Therefore, $|M|_{\text {max }}=\left|M_{1}\right|+\left|M_{2}\right|+\left|M_{3}\right|+\left|M_{4}\right|=414$.