Answer : The only numbers that work are multiplies of $10$ , multiplies of $2$ but not multiplies of $5$ such that $v_2 (n) \geq 4$ and multiplies of $5$ but not multiplies of $2$ such that $v_5(n) \geq 2$ Multiplies of $10$ trivially work. For $\gcd(n,10)=1$ we can give the construction $x=1111\dots 11=\frac {10^k-1}{9}$, taking $k$ a multiply of order of $10 \pmod n$ such that $v_3(x) \geq 3$ we have that $x \equiv 0 \pmod n$, so no such number exist. For $\gcd(10,n)=5$ such that $v_5(n) =1$ , take $n=5m$ we have $\gcd(m,10)=1$ and we can take the number $x=5555\dots 5=5\cdot \frac{10^k-1}{9}$ and doing the same as above we can get a $k$ such that $x \equiv 0 \pmod n$. For $\gcd(10,n)=5$ such that $v_5(n) \geq 2$ , assume there exists an $x=aaaa \dots aa=a \cdot \frac{10^k-1}{9}$ such that $x\equiv 0 \pmod n$ clearly such number is never divisible by $25$ ,contradiction , so these $n$ works. For $\gcd(10,n)=2$ such that $v_2(n) \leq 3$ we take $x=88888\dots 888 =8 \cdot \frac{10^k-1}{9}$ and we can get a $k$ such that $x \equiv 0 \pmod n$ For $\gcd(10,n)=2$ such that $v_2(n) \geq 4$ , assume there exists $x=aaa\dots aa=a \cdot \frac{10^k-1}{9}$ such that $x\equiv 0 \pmod n$ but $x$ is never divisible by $16$ , contradiction , so these $n$ work too. $\blacksquare$