How can a filter have zero group delay?
There are a couple of interesting aspects of "reconstruction to unity". First, there are two ways of combining two filters: parallel and in series. For a parallel topology it is ALWAYS possible to find a complimentary filter so that the pairs add to unity. It's easy enough, actually. Simply do $\tilde{H}(\omega) = 1-H(\omega)$. In the time domain that means that the impulse response of the complimentary filter is simply the negative of the original impulse response with 1 added to the first sample. So all the "ringy" stuff cancels out. Now the shape of this complimentary filter is not always what one would expect. For a 1st order low pass it actually is a first order high pass but for higher order filters it tends to have over/under swings in the cutoff region. However, it always exists as a stable causal filter.

Series (or cascade) "reconstruction to unity" is bit more complicated. Obviously the filters would have to be the inverse of each others, i.e. $\tilde{H}(\omega) = \frac{1}{H(\omega)}$. In general this can be done for any minimum phase filter. The inverse of a minimum phase filter is minimum phase as well and both are causal and stable. 

So this leaves us with the question of how to interpret group delay in these cases. The cascade case is actually the more interesting. Since the filters are inverse of each other, the phase, and hence the group delay, of one is the negative of the other. So at frequencies were one filter has positive group delay, the other has negative group delay. An easy example would be a low shelf with +6dB of gain and a low shelf with 6dB of cut.
So negative group delays are very real and certainly not a violation of causality. In practice, these show up in areas of the filter that are fairly "non-flat" so the traditional interpretation of "delay of the envelope" doesn't quite apply since there is a fair amount of amplitude distortion as well.

If you Google "negative group delay", you can find a few IEEE articles that have tackled the subject.