You have to be more abstract, John.  Finding patterns is useful, but most of the time is hopeless.  But abstract patterns are much easier to find.

If a number n is special, then one of the following must be true:

n = k
n = k + (k + 1)
n = k + (k+ 1) + (k + 2)
n = k + (k+ 1) + (k + 2) + ...

Where k <= n.

So expanding on this, we get:

n = k or n = 2k + 1 or n = 3k + 6 or n = 4k + 10 or ...

And so on for some positive integer k.  First thing to notice is that all the odds other than 1 are covered because every odd number can be expressed as 2k + 1.  So all we have left are the evens.

We can continue that pattern, getting more and more numbers, but it will get us nowhere (or so I think).  We won't ever be able to say we have all the numbers.

So lets do the same thing, but write numbers a different way:

n = k
n = k-1 + k + k+1
n = k-2 + k-1 + k + k+1 + k+2
n = k-a +... + k-1 + k + k+1 + ... + k+a

For some integers k or a.  But we find something weird happens when we add those up;

n = k-1 + k + k+1 = 3k
n = k-2 + k-1 + k + k+1 + k+2 = 5k
n = k-a +... + k-1 + k + k+1 + ... + k+a = ak

So from this fact, we can construct a proof that if n is divisble by an odd number, it must be special.

Now if you know a little (but very important) fact, you can figure out where to go from here.  But assuming you don't, that little fact is:

Every number can be written as a product of primes.

The only numbers that can be written as solely a product of 2's is a power of 2.  Thus, for any non-power of two, there must be a prime in that product that is not 2.  So it must be the case that...

... such a number is divisble by an odd number.

And that concludes that half of the proof (informally).  The other half can be deduced by finding that any special number must contain a prime factor other than 2, and every power of two must contain only powers of 2, so they can't be equal.

Right, I realized this when I wrote it out on paper.  However, you will notice that:

(-1 + 0 + 1) + 2 + 3 + 4 + 5 = 2 + 3 + 4 + 5, and thus, 14 is special.

It takes a little fix, but you find that if k - a < 0, then k-a + ... + a-k = 0, and there must be at least two terms to the right of a-k since k >= 1 and (k+a) - (a-k) = 2k.