Suppose that exist a $k$ such that all the multiple of k miss some digit from ${0,1,2,...,9}$ 1.Easy to see that $A=\sum^n \frac{1}{n}=\frac{1}{1}+\frac{1}{2}+...=$ infinity and A is also the sum of $k$ multiple inverse.It is diverge. 2.In otherway We will prove that the sum of k-multiples inverse missing at least one digit from ${0,1,2,...,9}=B$ is converge. Each multiple of k with n digits is $<10^n$ and we have not great than $9^{n-1}$ multiple of k with n digits missing at least one digit so $B<\sum^n{\frac{9^{n-1}}{10^n}}=9.(\frac{9}{10})^n$ which is coverge. So we get the contracdiction.

I didnt understand your solution. (I understood what you proved, but I dont think it solves the problem). Could anyone explain or complete his solution please?

We will construct $k$ such that $ak$ has all digits for all $1 \le a \le n$. In the following, $[0]$ denotes a block of $n$ digits, all zeroes. Take \[{k = 1[0]2[0]3[0]4[0]5[0]6[0]7[0]8[0]9[0]16[0]32[0]\cdots[0](2^{20} \mod 100)[0]25[0]125[0]\cdots [0](5^{12} \mod 10^5)}\] If $(a, 10) = 1$, we are done (just look at the 1,2,...,9 in the beginning, their multiples mod 10 are just a permutation of the digits). WLOG $10 \nmid a.$ If $a$ is a power of two, note that the sequence $2^k \mod 100$ has all digits, and we are done, since the blocks will cycle mod 100. If $a$ is a power of five the same argument works for the other blocks, so done.