6. If $n \in\{1,2, \cdots, 100\}$, and $n$ is a multiple of the sum of its digits, then there are $\qquad$ such $n$.

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6.33.

Let $n=\overline{a b}=10 a+b,(a+b) \mid (10 a+b)$.
When $a, b$ is 0, it obviously meets the requirement. Therefore, all one-digit numbers and two-digit numbers ending in 0, as well as the three-digit number 100, meet the requirement, making a total of 19 such $n$.

When $a, b$ are both not 0, since $\frac{10 a+b}{a+b}=\frac{9 a}{a+b}+1$, then $\frac{9 a}{a+b}$ must be an integer. If $(a+b, 3)=1$, it would lead to $(a+b) \mid a$, which is a contradiction. Therefore, $3 \mid (a+b)$. At this time, $a+b \in \{3,6,9,12,15,18\}$.

If $a+b$ is even, then $9 a$ is even, so $a, b$ must be even. Therefore, when $a+b=6$, only 42 and 24 meet the requirement; when $a+b=12$, only 48 and 84 meet the requirement; when $a+b=18$, there is no solution.

If $a+b$ is odd, when $a+b=3$, only 12 and 21 meet the requirement; when $a+b=9$, the solutions where $a, b$ are not 0 are 18, 27, ..., 81, a total of 8; when $a+b=15$, from $\frac{9 a}{a+b}=\frac{3 a}{5}$, we get $a=5$, then $b=10$, which is a contradiction.
Therefore, such $n$ totals $19+2+2+2+8=33$.