Parity of Sum of $n$ Consecutive Positive Integers

First, consider the case where $n$ is odd. A sequence of $n$ consecutive integers can have either $\frac{n-1}{2}$ or $\frac{n+1}{2}$ odd integers. If $n\equiv1\bmod{4}$ , then $\frac{n-1}{2}$ is even, and if $n\equiv-1\bmod{4}$ , then $\frac{n+1}{2}$ is even. If the number of odd integers in the sum is even, then the sum itself is even, so all odd $n$ are to be included.

Now consider even $n$ . In the sequence of $n$ consecutive integers, we will always have exactly $\frac{n}{2}$ odd integers. If $n\equiv0\bmod{4}$ , then $\frac{n}{2}$ is even and the sum will always be even. However, if $n\equiv2\bmod{4}$ , then $\frac{n}{2}$ is odd, and the sum will always be odd, as it contains an odd number of odd integers. Thus, the only integers NOT to be included in our count are those where $n\equiv2\bmod{4}$ , of which there are $504$ less than $2017$ (this can be calculated as $\frac{2014-2}{4}+1$ ), leaving $2017-504=\boxed{1513}$ integers such that there is at least one sequence of $n$ consecutive positive integers with even sum.