Group delay of the FIR filter
To be precise the group delay of a linear phase FIR filter is $(N-1)/2$ samples, where $N$ is the filter length (i.e. the number of taps). The group delay is constant for all frequencies, because the filter has a linear phase, i.e. its impulse response is symmetrical (or asymmetric). A linear phase means that all frequency components of the input signal experience the same delay, i.e. there are no phase distortions. So for a frequency selective filter (e.g., a low pass filter), if the input signal is in the passband of the filter, the output signal is approximately equal to the input signal delayed by the group delay of the filter. Note that in general FIR filters do not have a linear phase response. In this case, the group delay is a function of frequency.

EDIT:

Consider an odd length FIR filter with even symmetry ("type I filter"). In this case the group delay is an integer number of samples $K=(N-1)/2$. The filter has a linear phase response (because the group delay is the negative derivative of the phase) and the filter's frequency response can be written as

$$H(\omega)=A(\omega)e^{-jK\omega}$$

where $A(\omega)$ is the real-valued amplitude function, and $\phi(\omega)=-K\omega$ is the (linear) phase. If the input signal's Fourier transform is $X(\omega)$, then the Fourier transform of the output signal is given by

$$Y(\omega)=H(\omega)X(\omega)=A(\omega)X(\omega)e^{-jK\omega}\tag{1}$$

Now assume that the frequency content of the input signal is in the passband of the filter, i.e. in the frequency region where $A(\omega)\approx 1$ holds. Then from (1) the output signal's spectrum can be written as

$$Y(\omega)\approx X(\omega)e^{-jK\omega}\tag{2}$$

and, from the shift property of the Fourier transform, the output signal is (approximately) given by

$$y(n)\approx x(n-K)$$

i.e. $y(n)$ is a delayed version of the input signal, where the delay is given by the group delay $K$.