For two-digit $n=10a+b$, we have $6(10a+b)=100a+10c+b\;\forall a,b,c \in \mathbb{N}, 0 \le a,b,c \le 9$. This gives the trivial solution of $n=18$. Now we prove that it is the only one; there are no other larger numbers. This is quite evident when we see that $6$ is too small; to get a four-digit number from a three-digit number by multiplying by $6$ requires the number to be at least $167$, after we see that the second digit cannot be $6$ when trying numbers in the range $\lbrace 167,200 \rbrace$. A similar argument suffices for the range $\lbrace 200,300 \rbrace$, after which we see that $\lbrace 300,999 \rbrace$ cannot work. This argument extends as $n \rightarrow \infty$, so we are done. $\blacksquare$