Integers*?
Let [tex]n[/tex] be the smallest of these five consecutive integers. Then the sum of the five consecutive integers is [tex]\frac{5(2n+4)}{2}=5n+10,[/tex] so we seek the largest positive integer [tex]n[/tex] such that [tex]5n+10<2020.[/tex] Solving this inequality, we find that [tex]n<402,[/tex] so there are [tex]\boxed{401}[/tex] such positive integers.
