Suppose n is an odd number. Then, by definition, n=2k+1, for some integer k.
But then n = (k)+(k+1), so n is the sum of two consecutive integers. If n>1, then k>0, so n is the sum of two positive consecutive integers.
Now suppose that n is an even number, but that n/2 is odd, and n/2>1. Another way to say that is, suppose n is divisible by 2, but is not divisible by 4, and ``n>2
Then: n/2 = k+(k+1) (we know n/2 is odd and greater than 1, and we already figured out we could write an odd number as the sum of two consecutive positive integers)
So: n = k + k + (k+1) + (k+1)
subtract 1 from the first term and add it to the last term, and you get:
n= (k-1) + k + (k+1) + (k+2)
Thus n can be written as the sum of four consecutive integers.
n>2 and n is even, so n is at least 4. But n is not divisible by 4, so n is at least 6. So k is at least 1. If k>1, then all of the numbers are positive, and n is the sum of four consecutive positive integers. If k=1, then the first term is 0, but then we may ignore it, and n is the sum of three consecutive positive integers.
Now suppose that n is even, and is divisible by 4, but it is not divisible by 8. That means n/2 is even but is not divisible by 4. But we already know that numbers that are even but not divisible by 4 can be written as the sum of four consecutive integers:
n/2 = (k-1) + k + (k+1) + (k+2)
So: n = (k-1) + (k-1) + k + k + (k+1) + (k+1) + (k+2) + (k+2)
subtract 2 from the first (k-1) and add it to the last (k+2)
subtract 1 from the second (k-1) and add it to the first (k+2)
subtract 1 from the first k and add it to the second (k+1)
n = (k-3) + (k-2) + (k-1) + k + (k+1) + (k+2) + (k+3) + (k+4)
Thus a number that is divisible by 4 but not by 8 may be written as the sum of eight consecutive integers.
(again, you have to be a little careful to show that you have positive numbers, except possibly the first term is 0)
this pattern will continue
(your profile says 9th grade, so my guess is you aren't expected to use induction)
The only numbers that don't get taken care of are the powers of 2. Every other positive integer may be divided repeatedly by 2, and eventually you will wind up with an odd number that is at least 3.
If you divide a power of 2 repeatedly by 2, you will eventually wind up with the odd number 1, and 1 can't be written as a sum of consecutive positive integers, so the argument breaks down.