, 2021

Solution:
One can show that if $m>0$, then we must either have $n>3003$ or $n<1001$. If $n<1001$, each term other than $n-1001$ has absolute value at least $1000$, so $m>1000^{5}$, meaning that $m$ has at least $16$ digits. However, if $n>3003$, it is clear that the minimal $m$ is achieved at $n=3004$, which makes
$$m=2002 \cdot 1002 \cdot 1001 \cdot 3 \cdot 2 \cdot 1=12 \cdot 1001 \cdot 1001 \cdot 1002$$
which is about $12 \cdot 10^{9}$ and thus has $11$ digits.