Sum of certain consecutive numbers is $1000$. Question:

The sum of a certain number (say $n$) of consecutive positive integers is $1000$. Find these integers.

I have no idea how to approach the problem. I did try the following but did not arrive anywhere: I said that
$$1+2+3+....+44=990$$
Then I subtracted numbers from $1$ to $9$ and added $45$. Then subtracted $10$ and added $46$ and continued the process. But arrived nowhere.
Thanks for the help!!
P.S. I do know that $n=1$ such that the "numbers" belong to the set $S=\{1000\}$ is trivial solution but looking for others. 
Edit : I am asking for the integers and not the number of ways it can be done in. 
 A: You could think as follows:
Consecutive numbers:
$$
\ldots, r-2, r-1, r, r+1, r+2, \ldots
$$
If you sum these up you will get 
$$
n\cdot r
$$
Where $n$ is an odd number
So we want $n\cdot r= 1000$
Try for example $n=5$
hope this helps :)
Edit: read the comments below 
A: The sum you're looking at is
$$
a+(a+1)+\dots+(a+n-1)=na+(1+2+\dots+(n-1))=na+\frac{n(n-1)}{2}
$$
so you get the equation
$$
n^2+(2a-1)n-2000=0
$$
An integer solution has to be a divisor of $2000$, so of the form $n=2^x5^y$, with $0\le x\le 4$ and $0\le y\le 3$. The condition is then that
$$
2a-1=\frac{2000-n^2}{n}=\frac{2000}{n}-n
$$
This number must be odd. If $n$ is even (that is, $x>0$), we need that $2000/n$ is odd, so $x=4$. If $n$ is odd (that is, $x=0$), any value of $y$ is good.
Thus we have $n\in\{1,5,25,125,16,80,400,2000\}$.
If you want $a>0$, then $n^2<2000$ and the choices are reduced to $n\in\{1,5,25,16\}$, corresponding to
\begin{array}{cc}
n & a \\ \hline
1 & 1000 \\
5 & 198 \\
25 & 28 \\
16 & 55
\end{array}
A: The sum of $n$ consecutive numbers is $n$ times the average. I.e.
$$n\frac{i+i+n-1}2=\frac{n(2i+n-1)}2.$$
We can look for $n$ among the factors of $2000$, using
$$i=\frac12\left(\frac{2000}n-n+1\right).$$
The ratio $$\frac{2000}n$$ must have the opposite parity of $n$.
