19. $\mathrm{n}$ is a positive integer no greater than 100 and no less than 10, and $\mathrm{n}$ is a multiple of the sum of its digits, such $\mathrm{n}$ has $(\quad)$ numbers.
【Answer】 24
【Analysis】(Method one) Construction method, given $10 \leq n \leq 100$, when $\mathrm{n}=100$, it clearly fits. The rest of the $\mathrm{n}$ are two-digit numbers. The following is to find the number of two-digit $\mathrm{n}$.

Let $n=\overline{a b}=10 a+b(a, b$ are integers and $a \geq 1)$; we get $(a+b) \mid(10 a+b)$;
Construct $10 a+b=10(a+b)-9 b$, and find that $9 b$ is an integer of $(a+b)$.

Classify $\mathrm{b}$: (1) When $b=0$, $a=1,2,3,4,5,6,7,8,9$ for a total of 9 cases;
(2) When $b=1$, $a=2,8$ for a total of 2 cases;
(3) When $b=2$, $a=1,4,7$ for a total of 3 cases;
(4) When $b=3$, $a=6$ for a total of 1 case;
(5) When $b=4$, $a=2,5,8$ for a total of 3 cases;
(6) When $b=5$, $a=4$ for a total of 1 case;
(7) When $b=6$, $a=3$ for a total of 1 case;
(8) When $b=7$, $a=2$ for a total of 1 case;
(9) When $b=8$, $a=1,4$ for a total of 2 cases;
(10) When $b=9$, there is no solution for $a$. There are $9+2+3+1+3+1+1+1+2=23$ two-digit numbers that fit. In summary, there is also 100, which is 1 more, making a total of $23+1=24$.
(Method two) Direct enumeration.
Tens place is 1, there are $10,12,18 ; 3$ cases
Tens place is 2, there are $20,21,24,27 ; 4$ cases
Tens place is 3, there are 30,$36 ; 2$ cases
Tens place is 4, there are $40,42,45,48 ; 4$ cases
Tens place is 5, there are 50,$54 ; 2$ cases
Tens place is 6, there are 60,$63 ; 2$ cases
Tens place is 7, there are 70,$72 ; 2$ cases
Tens place is 8, there are $80,81,84 ; 3$ cases
Tens place is 9, there are $90 ; 1$ case
Add 100, which is 1 more; a total of 24.