let $n$ be such a positive integer, and let $S$ be the set of its digits. $0 \notin S$, as no positive integer is a multiple of $0$. note that $S$ must contain an even number, since otherwise it would have at most five elements. this implies $5 \notin S$, since if it were, $n$ would be a multiple of both $5$ and some even number, and hence end in a $0$, which is bad. if $9$ were not in $S$, then the sum of the digits of $n$ would be $1 + 2 + 3 + 4 + 6 + 7 + 8 = 31$, which is not a multiple of $9$, implying $9 \nmid n$, a contradiction. hence $9 \in S$, implying $9 \mid \textstyle\sum S$. since $1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 = 40$, we must leave out $4$ in order to attain a digit sum divisible by $9$. thus, $S = \{1, 2, 3, 6, 7, 8, 9\}$.