First, the number must be a multiple of 9 and 7. The first is easy to check and holds for all permutations. Note that when two adjacent digits $a$ and $b$ are swapped, the number changes by $9(a-b) \cdot 10^{k}$ (we disregard sign), so $9(a-b)$ must also be a multiple of 63 for all digits $a$ and $b$. In particular, this is sufficient, since a permutation can be represented as a series of transpositions. This means that $a-b$ must be a multiple of 7 for all digits $a$ and $b$, so either all digits are equal or they are in $\{0,7\},\{1,8\}$, or $\{2,9\}$. We find the minimum for each case separately. We first provide the following useful fact: the first repunit (numbers $1,11,111, \ldots$) that is a multiple of 7 is 111111. This is because $10 \bmod 7=3$, and 3 is a generator modulo 7 (of course, you can just compute the powers of 3 by hand, and it will not take much longer). If a number $k \cdot 1 \ldots 1$ is a multiple of 63, then either $k$ or $1 \ldots 1$ is a multiple of 7; if it is $k$, then it's clear that we need 777777777 to make the sum a multiple of 9. If $1 \ldots 1$ is a multiple of 7, then it is at least 111111, then to make a multiple of 9, we need 333333. If the only digits are 7 and 0, then we need at least nine sevens to make the digit sum a multiple of nine, which has more digits than 333333. If the only digits are 8 and 1, then we can note that since 8 and 1 are both $1(\bmod 7)$, these numbers are equivalent to the repunits modulo 7, so such numbers have at least six digits. The best such six-digit number with digits summing to a multiple of 9 is 111888, which is our new candidate. If the only digits are 9 and 2, then by analogous logic such numbers have at least six digits. But the smallest such number is 999999, which is not better. So our best answer is 111888. It works.