How do you pick the wavenumber at which the group velocitiy is evaluated? The equation for group velocity is 
$
v_g(k) = \frac{d\omega}{dk}.$ This is obviously a function of $k$ but typically the word is used as if there is a single group velocity and not a whole function. How do you assign a group velocity to a "group" ? Is there any clear cut mathematical definition which gives another condition or is there a special $\omega$ at which the function is typically evaluated ? Or is the word a misnomer and there isn't a single group velocity which can be assigned to a wavepacket ?
 A: It depends a bit on context. For transmission of signals in optical cables, the bandwidth is very small compared to the frequencies of the laser modes. So the meaning of group velocity is unambiguous. (But over a long distance, there will be noticeable dispersion, which limits the data rate.)
In a case like water waves, the situation is different. But there one can say that for deep water the group velocity is half the phase velocity. Both depend on wavelength: 
$$c_\phi = \sqrt{\frac{g\lambda}{2\pi}} = \frac{g}{2\pi}T.$$
A: Typically the "group" is a localized pulse in space.  When you look at the k spectrum, it is typically also localized in k-space, with some average value $k_0$.  (For a Gaussian wave packet, for example, the k spectrum, or Fourier transform, is also a Gaussian, and $k_0$ is the value where it peaks.)  The group velocity then corresponds to $\text{d} \omega\,/\text{d} k$ evaluated at $k_0$.  The choice is only ambiguous if the k-space distribution is such that the average value of k is undefined, though it is also helpful if the Taylor series expansion for $\omega(k)$ about $k_0$ converges quickly.  If not, then the wave packet will tend to change shape as it propagates, which makes the idea of a group velocity fuzzier.
UPDATE with more details:
Suppose your wave packet is given by
$\psi(x,t) = \int\limits_{-\infty}^\infty A(k) e^{i(kx-\omega t)} \mathrm{d} k$.
Let the average wave number be $k_0$ (sorry for the notation conflict with OP's $k_0$), and expand your expression for $\omega(k)$ in a Taylor series about $k_0$, so that
$\omega(k) = \omega(k_0) + \left(\dfrac{\partial \omega}{\partial k} \right)_{k = k_0} \left(k - k_0 \right) + \mathcal{O}((k - k_0)^2)$.
As long as the range of relevant $k$ values is small enough that we can throw away the 2nd order and higher terms, we can use this to write
$kx - \omega t \approx (k - k_0) x - \left(\dfrac{\partial \omega}{\partial k} \right)_{k = k_0}\left( k - k_0\right) t + k_0 x - \omega(k_0) t  $, 
so that $\psi(x,t)$ becomes
$\psi(x,t) \approx \int\limits_{-\infty}^\infty A(k) \exp\left[i (k - k_0)\left(x - \left(\dfrac{\partial \omega}{\partial k} \right)_{k = k_0} t \right) \right]
e^{i(k_0 x - \omega(k_0) t)} \mathrm{d} k $.
Inspecting this form, we see that it involves a plane wave traveling at the phase speed $\omega(k_0)/k_0$ (that's the last exponential factor, which comes out of the integral), modulated by an envelope traveling at the group speed $\left(\dfrac{\partial \omega}{\partial k} \right)_{k = k_0}$ (from the "x - vt" factor in the square brackets).  So, as long as the 1st order Taylor series is a good approximation, the group velocity is $\partial \omega / \partial k$ computed at the average value of $k$.
