smc

We might at first think that the answer would be $9$, because $1+2+3\cdots +n=45$ 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, $\boxed{90}$ 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.