Iranian Mathematical Olympiad

We claim that $N = (1389!)^{m+1} - 1$ has the desired properties. Let $2 \le b \le 1389$, we know that there are at least $m+1$ zeros in the rightmost of $(1389!)^{m+1}$ in base $b$, because $(1389!)^{m+1}$ is divisible by $b^{m+1}$. Hence, the $m+1$ rightmost digits of $(1389!)^{m+1} - 1$ is $b-1$, then sum of the digits of $N$ in base $b$ is at least $(m+1)(b-1)$ which is greater than or equal to $m+1$.