Half of the FFT transform from real data is redundant. However, the imaginary parts of bin 1 and bin 513 are zero, so that's where the redundancy in those bins is. As a result, you need bins 1 to 513 to represent the original data (you could stuff the real part of bin 513 into the imaginary part of bin 1 if you really wanted to get along with 1024 words of storage).
Your question is whether bin 513 is useful. That depends on what you are doing. It is quite unlikely that it is significantly less useful than bin 512, however.
If you are using FFT as a mechanism for implementing convolution, you won't want to lose information, so you'll need bin 513.
If you are using it as an analysis tool, it is likely that you will be working with analog prefiltering and windowing and other tools that make higher frequencies taper off. But then the tapering off will not just hit bin 513 but will affect quite a number of bins before that as well.
On its own, the FFT/DFT is a representation of the frequencies of a periodic signal with a period of 1024 (in your case). But it's rarely data from an actual circle that you are applying it on: FFT is almost always applied as a component in some more complex algorithm, using padding, windowing, overlap-add and other stuff.
What the bins mean and how important retaining every one of them is depends on just what you use the FFT for.