Theoretical question on Fast Fourier
This is (trivially) possible because all Fourier transform like operations are ideally invertible. So what you ask for is actually easily realized by:

$X(e^{j\Omega}) = DTFT\{ IDFT\{X(k)\} \}$

In formulae this is 

$X(e^{j\Omega}) = \sum_{l=0}^{N-1} \sum_{k=0}^{N-1} X(k)e^{2 \pi j kl/N} e^{-j\Omega l}$

which can be a bit simplified to

$X(e^{j\Omega}) = \sum_{k=0}^{N-1} X(k) \sum_{l=0}^{N-1} e^{jl (2 \pi k/N-\Omega)}$

or, finally

$X(e^{j\Omega}) = \sum_{k=0}^{N-1} X(k) \frac{e^{2 \pi j k- jN\Omega}-1}{e^{2 \pi j k/N-\Omega}-1}$.

EDIT: The question on whether you need all samples X(k) lies in the behavior of the interpolation function $I(x) = \frac{e^{j N x}-1}{e^{j x}-1}$. This is because the above result is then equal to

$X(e^{j\Omega}) = \sum_{k=0}^{N-1} X(k) \, I(2 \pi k/N - \Omega)$.

I(x) has some "sinc like" behavior, so the question on how many X(k) you need, is really when I(x) has decayed to a value "small enough".