Let [tex]x[/tex] be the smallest of the consecutive integers summing to 2020. Then the largest integer in the sum is [tex]x+4039,[/tex] so our answer is [tex]\boxed{4039}.[/tex]
For completeness, though, let's solve for the value of [tex]x[/tex] that satisfies the conditions of the problem. Recall that the sum of an arithmetic series with first term [tex]a_1,[/tex] last term [tex]a_n,[/tex] and [tex]n[/tex] terms, is [tex]\frac{n(a_1+a_n)}{2}.[/tex] Plugging in our known values gives us the equation [tex]\frac{4040(2x+4039)}{2}=2020,[/tex] which simplifies to [tex]2x+4039=1.[/tex] Thus, [tex]x=-2019,[/tex] so the 4040 consecutive integers are [tex]-2019,-2018,\dots,-1,0,1,\dots,2018,2019,2020.[/tex] Notice how everything except the [tex]2020[/tex] cancels out!
