Numbers! no less than a maze

Suppose $n<1000$ was a $3$ -digits number

$n=\overline{abc}$

To respect the constraint we need to have

$100a+10b+c=6(a+b+c)\iff 94a+4b=5c$

Since $a$ , $b$ and $c$ are digits and $a$ is non-zero:

$94a+4b\geq 94\cdot 1 +4\cdot 0\geq 94$ and

$5c\leq 5\cdot 9 \leq 45$

Which is not possible, so $n$ has less than $3$ digits

Suppose $n=\overline{ab}$

$\Longrightarrow 10a+b=6a+6b\iff 4a=5b \iff a=5,b=4$

$n$ cannot have only one digit, since that would imply

$a=6a \iff a=0$

So there is only $\boxed{1}$ such $n$

$\textbf{Note:}$

You should include that we are talking about positive integers, otherwise $0$ would indeed be a solution