Why is a linear phase important?
Let me add the following graphic to the great answers already given, with the intention of a specific and clear answer to the question posed. The other answers detail what linear phase is, this details why it is important in one graphic:

When a filter has linear phase, then all the frequencies within that signal will be delayed the same amount in time (as described mathematically in Fat32's answer).
Any signal can be decomposed (via Fourier Series) into separate frequency components. When the signal gets delayed through any channel (such as a filter), as long as all of those frequency components get delayed the same amount, the same signal (signal of interest, within the passband of the channel) will be recreated after the delay.
Consider a square wave, which through the Fourier Series Expansion is shown to be made up of an infinite number of odd harmonic frequencies.
In the graphic above I show the summation of the first three components. If these components are all delayed the same amount, the waveform of interest is intact when these components are summed. However, significant group delay distortion will result if each frequency component gets delayed a different amount in time.
The following may help give additional intuitive insight for those with some RF or analog background.
Consider an ideal lossless broadband delay line (such as approximated by a length of coaxial cable), which can pass wideband signals without distortion.
The transfer function of such a cable is shown in the graphic below, having a magnitude of 1 for all frequencies (given it is lossless) and a phase negatively increasing in direct linear proportion to frequency. The longer the cable, the steeper the slope of the phase, but in all cases &quot;linear phase&quot;.
This makes sense; the phase delay of 1 Hz signal passing through a cable with a 1 second delay will be 360°, while a 2 Hz signal with the same delay will be 720°, etc...
Bringing this back to the digital world, $z^{-1}$ is the z-transform of a 1 sample delay (therefore a delay line), with a similar frequency response to what is shown, just in terms of H(z); a constant magnitude = 1 and a phase that goes linearly from $0$ to $-2\pi$ from f = 0 Hz to f = fs (the sampling rate).

The simplest mathematical explanation is that the a phase that is linear with frequency and a constant delay are Fourier Transform pairs. This is the shift property of the Fourier Transform. A constant time delay in time of $\tau$ seconds results in a linear phase in frequency $-\omega \tau$, where $\omega$ is the angular frequency axis in radians/sec:
$$\mathscr{F}\{g(t-\tau)\} = \int_{-\infty}^{\infty}g(t-\tau)e^{j\omega t}dt$$
$$u = t - \tau$$
$$\mathscr{F}\{g(u)\} = \int_{-\infty}^{\infty}g(u)e^{-j\omega (u+\tau)}du$$
$$ = e^{-j\omega \tau}\int_{-\infty}^{\infty}g(u)e^{-j\omega u}du$$
$$ = e^{-j\omega \tau}G(j\omega)$$