Solution:
Yes. In general, let $N$ be a number of the form $N=2^{k} p$, where $p$ is an odd prime less than $2^{k+1}$. We will show that one can form an expression, obtained by adding and subtracting together all the divisors of $N$, which is equal to zero. First note that
$$2^{k} p-2^{k-1} p-\cdots-p=p .$$
The remaining divisors are $1,2, \ldots, 2^{k}$, whose sum is more than $p$. Thus we are able to write $\left(2^{k+1}-1+p\right) / 2$ as a sum of some subset of the remaining divisors simply by considering its binary representation. Clearly the unused terms will sum to $\left(2^{k+1}-1-p\right) / 2$. But the difference between these two quantities is $p$, just as above. It is now clear how to form the desired expression involving all divisors of $N$ which evaluates to zero, and you're done.

If $N$ is to be stable, then the factors of $N$ can be put on either side of an equal sign, and when these factors are summed, you get an equality. The highest power of $2$ that divides $N$ will either lie on the same side as $N$, or the opposite side. The key idea of David's solution is the observation that if $N$ is stable and the highest power of $2$ that divides $N$ lies on the opposite side of the equal sign from $N$, then $2N$ is also stable.
For example, $12$ is stable, because
$$1+3+4+6=2+12 \text{,}$$
and notice that $4$ lies opposite $12$. To show that $24$ is stable, we let $4$ and $12$ exchange places, and then add in the two new factors ($8$ and $24$), getting
$$1+3+12+6+8=4+24 \text{.}$$
It is easy to see why this method works (verify it!)
The remainder of David's argument looks at $2008$, which is not stable, but he successively doubles it, making it more and more balanced, until he shows that $2008 \cdot 2^{4}$ is stable. Then the algorithm described above will show that $2008 \cdot 2^{k}$ is stable for all $k \geq 4$.

Solution:
(Sketch) This solution, by David Spies of Albany High School, won the Brilliancy Award, because it used ideas that no other solution had.
If $N$ is to be stable, then the factors of $N$ can be put on either side of an equal sign, and when these factors are summed, you get an equality. The highest power of $2$ that divides $N$ will either lie on the same side as $N$, or the opposite side. The key idea of David's solution is the observation that if $N$ is stable and the highest power of $2$ that divides $N$ lies on the opposite side of the equal sign from $N$, then $2N$ is also stable.