2019 AMC 10A 第 5 题

A long consecutive block can be centered around zero

Use divisibility of $90$ by the number of terms

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.