What is the greatest number of consecutive integers whose sum is $45?$
$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$
We might at first think that the answer would be $9$, because $1+2+3 \dots +n = 45$ when $n = 9$. But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence $-44, -43, \cdots, 44, 45$ cancels out except $45$.  Thus, the answer is, intuitively, $\boxed{\textbf{(D) } 90 }$ integers.
Though impractical, a proof of maximality can proceed as follows:  Let the desired sequence of consecutive integers be $a, a+1, \cdots, a+(N-1)$, where there are $N$ terms, and we want to maximize $N$.  Then the sum of the terms in this sequence is $aN + \frac{(N-1)(N)}{2}=45$.  Rearranging and factoring, this reduces to $N(2a+N-1) = 90$.  Since $N$ must divide $90$, and we know that $90$ is an attainable value of the sum, $90$ must be the maximum.