for his correct solution!
Here is a suggested solution by other:
Suppose $n$ has digits $a_1a_2\cdots a_k$ and digit sum $s=\sum a_i$. If we temporarily allow digits larger than 9, then $4n=(4a_1)(4a_2)\cdots(4a_k)$. Each carry of one reduces the digit sum by 9, so after making all necessary carries, the digit sum for $4n$ is at most $4s$. It can only be $4s$ iff there are no carries. Similarly, $11n$ has digit sum at most $2s$, with equality iff there are no carries.
Since $44n$ has digit sum $8s$, there cannot be any carries. In particular, there are no carriers in forming $4n$, so each digit of $n$ is at most 2. Hence, there are no carries in forming $3n$ and the digit sum of $3n$ is 300.