Every positive integer divide some number whose representation (base 10) contains only zeroes and ones.
One can prove that:
Consider the numbers 1, 11, 111,
I think this is a fair and interesting question.
Please note that though what you describe is a proof there always exist such number, the found number will not always be minimal.
Only solution I can think of is to compute the remainders of the powers of 10 modulus the given n and than try to construct a sum giving remainder 0 modulo n using at most one of each of these powers. You will never need more than n different powers(which you prove i your question).