4. The integer $n$ satisfies the inequality $n+(n+1)+(n+2)+\cdots+(n+20)>2019$. What is the minimum possible value of $n$ ?
SolUtion
087

We solve the inequality $n+(n+1)+(n+2)+\ldots+(n+20)>2019$ Therefore, $21 n+210>2019$, i.e. $21 n>1809$ and $7 n>603$. Therefore, $n>\frac{603}{7}=86.1 \ldots$, so $n$ must be at least 87 .