2019 AMC 10 A

We might at first think that the answer would be , because when . But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence , , , , cancels out except . Thus, the answer is, intuitively, integers.

Though impractical, a proof of maximality can proceed as follows: Let the desired sequence of consecutive integers be , , , , where there are terms, and we want to maximize . Then the sum of the terms in this sequence is .Rearranging and factoring, this reduces to . Since must divide , and we know that is an attainable value of the sum, must be the maximum.

To maximize the number of integers, we need to make the average of them as low as possible while still being positive. The average can be if the middle two numbers are and , so the answer is .