13. (12 points) If a positive integer $N$ satisfies that for every multiple $\overline{\mathrm{abc}}$ of $N$, $\overline{\mathrm{bca}}$ and $\overline{\mathrm{cab}}$ are also multiples of $N$ (where $a$, $b$, and $c$ are integers from 0 to 9, and it is agreed that 123 represents 123, 028 represents 28, and 007 represents 7), then $N$ is called a “perfect divisor” (for example, 9 is a “perfect divisor”). The number of such “perfect divisors” is $\qquad$.
【Solution】Solution: (1) If $a, b, c$ are not all equal, then $N|\overline{\mathrm{abc}}, N| \overline{\mathrm{bca}}, N \mid \overline{\mathrm{cab}}$, then $N \mid(\overline{\mathrm{abc}}+\overline{\mathrm{bca}}+\overline{\mathrm{cab}})$, i.e., $N \mid(a+b+c) \times 111$, if $N \mid(a+b+c)$ holds for every $a, b, c$, then $N \mid 9$, then $N \mid 999$, i.e., $N$ is a divisor of 999, which are $1,3,9,27,37,111,333,999$, a total of 8, (2) If $a=b=c$, then the ratio is a perfect divisor, which are $111,222,333,444,555,666,777,888$, 999, a total of 9,

The ones that overlap with (1) are $111,333,999$, a total of 3,
Therefore, the "perfect divisors" are $8+9-3=14$ in total.
The answer is: 14.