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 ( ) numbers.
【Analysis】That is, $10 \leq n \leq 100$, when $\mathrm{n}$ is a three-digit number, $\mathrm{n}$ is 100, which meets the requirement. When $\mathrm{n}$ is a two-digit number, let's assume $n=\overline{a b}=10 a+b$, at this time $10 a+b$ is a multiple of $a+b$, so $a+b|10 a+b \Rightarrow a+b| 9 a$
One, $a+b=a$ that is $b=0$, then $a \mid 9 a$, at this time $a$ can take 1 to 9, a total of 9 kinds
When $b \neq 0$, since $a+b>a$, and $a+b \mid 9 a$, so at this time $a+b$ must be a multiple of 3
Two, $a+b=3$, $3 \mid 9 a$, because $b \neq 0$, so there are $12, 21$, 2 kinds
Three, $a+b=6$, $6 \mid 9 a$, so $a$ should be even, because $b \neq 0$, so there are $24, 42$, 2 kinds
Four, $a+b=9$, $9 \mid 9 a$, because $b \neq 0$, so there are $18, 27, \cdots \cdots, 81$, 8 kinds
Five, $a+b=12$, $12 \mid 9 a$, so $a$ should be a multiple of 4, because $b \neq 0$, so there are $48, 84$, 2 kinds
Six, $a+b=15$, $15 \mid 9 a$, so $a$ should be a multiple of 5, there is no number that meets the condition
Seven, $a+b=18$, $18 \mid 9 a$, so $a$ should be even, there is no number that meets the condition
In summary, there are $1+9+2+2+8+2=24$ numbers