Assume there is $n$ such that this does not hold. Now, we have 2 cases: Either $n$ and $3n$ have the same number of digits. Then the leftmost digit of $n$ can't be more than 3, thus it must be equal to 3. But then to leave the number of digits invariant $3n$ must have a digit 9 at this place. Contradiction! Either $3n$ has one digit more than $n$. Then this new digit must be the carry from the digit to the right of it after multiplication of 3. But this clearly must be either 1, either 2. Contradiction! In both cases, we obtain a contradiction and hence the assertion is proved.