2019 AMC 10A Problem 5

Suppose there are $k$ consecutive integers with first term $a.$ Their sum is $\dfrac{k(2a+k-1)}{2}=45,$ so $k$ must divide $90.$ Therefore the number of terms cannot exceed $90.$ This bound is attained by $$-44, -43, \cdots, 44, 45$$ which has $90$ terms and sum $45.$ Thus the greatest possible number of consecutive integers is $90.$ Thus, D is the correct answer.