What does it mean that the DFT equals the Complex Fourier Coefficients for even frequencies?
okay, i am using these definitions for the continuous Fourier Transform

$$ \mathscr{F}\Big\{x(t)\Big\} \triangleq X(f) \triangleq \int_{-\infty}^{+\infty} x(t) \, e^{-i 2 \pi f t} \, \mathrm{d}t  $$

and inverse

$$ \mathscr{F}^{-1}\Big\{X(f)\Big\} \triangleq x(t) = \int_{-\infty}^{+\infty} X(f) \, e^{i 2 \pi f t} \, \mathrm{d}f $$ 

$f$ is "ordinary" frequency, not angular frequency (which for the continuous-time signals, the notation we use is $\Omega \triangleq 2 \pi f$).  this is a preferred representation of the Fourier Transform for electrical engineers because it is the easiest manner to apply the Duality property (lotsa similarity between the FT and the inverse).

in this answer is shown how sampling in one domain (e.g. the "time" domain) causes periodic extension (repetition and overlapping and adding) in the reciprocal domain (e.g. the "frequency" domain).  this is pointed out without regard to bandlimiting, so if the input is not sufficiently bandlimited, there will be aliasing.  given the assumption of a real input to start with (which means the Fourier transform or two-sided spectrum has symmetry about $f=0$) frequency components above Nyquist (which is $\tfrac12$ the sample rate) are folded back to locations below Nyquist and added to whatever frequency components existed there.

now, your continuous-time input

$$ x(t) = \sum\limits_{k=-\infty}^{\infty} c_k \ e^{i 2 \pi k t} $$

has a spectrum (which is not necessarily bandlimited, and is actually not bandlimited since the $c_k$ coefficients are not zero above some finite $k$) of

$$ X(f) = \sum\limits_{k=-\infty}^{\infty} c_k \ \delta(f - k) $$

first defining the Dirac comb:

$$ \operatorname{III}_T(t) \triangleq \sum_{n=-\infty}^{\infty} \delta(t-nT) $$

now, the ideally sampled signal (still represented in the continuous-time domain) is 

$$\begin{align}
x_\text{s}(t) &amp;\triangleq x(t) \cdot T \cdot \operatorname{III}_T(t) \\
 &amp;= x(t) \cdot T \sum_{n=-\infty}^{\infty} \delta(t-nT) \\
 &amp;= T \sum_{n=-\infty}^{\infty} x(t) \ \delta(t-nT) \\
 &amp;= T \sum_{n=-\infty}^{\infty} x(nT) \ \delta(t-nT) \\
 &amp;= T \sum_{n=-\infty}^{\infty} x[n] \ \delta(t-nT) \\
\end{align}$$

and also this

$$\begin{align}
x_\text{s}(t) &amp;\triangleq x(t) \cdot T \cdot \operatorname{III}_T(t) \\
 &amp;= x(t) \cdot T \cdot \sum_{m=-\infty}^{\infty} \frac1T e^{i 2 \pi m \frac{t}{T}} \\
 &amp;= \sum_{m=-\infty}^{\infty} x(t) \ e^{i 2 \pi m \frac{t}{T}} \\
\end{align}$$

now from the latter, we know that the spectrum of the ideally sampling signal is this:

$$ \mathscr{F}\Big\{x_\text{s}(t)\Big\} = X_\text{s}(f) = \sum_{m=-\infty}^{\infty} X\left(f - \tfrac{m}{T} \right)
 $$

which is periodic in the frequency domain with period $\frac{1}{T}$ where your sampling period $T=\tfrac{1}{N}$.  that means your sample rate is $\tfrac{1}{T}=N$ and your Nyquist frequency is $\tfrac{N}{2}$.  that is the "folding frequency" or "foldover frequency". 

Okay, winding up, you plug the expression of $X(f)$, the non-bandlimited Fourier series, into the equation for $X_\text{s}(f)$, which is the spectrum of the ideally-sampled signal.  And you plug in $N$ for $\tfrac1T$.

$$ X(f) = \sum\limits_{k=-\infty}^{\infty} c_k \ \delta(f - k) $$

$$\begin{align}
 X_\text{s}(f) &amp;= \sum_{m=-\infty}^{\infty} X\left(f - \tfrac{m}{T} \right) \\
  &amp;= \sum_{m=-\infty}^{\infty} X(f - N m) \\
  &amp;= \sum_{m=-\infty}^{\infty} \sum\limits_{k=-\infty}^{\infty} c_k \ \delta\big((f - N m) - k\big) \\
  &amp;= \sum\limits_{k=-\infty}^{\infty} \sum_{m=-\infty}^{\infty} c_k \ \delta\big(f - (k+Nm) \big) \\
  &amp;= \sum\limits_{k=-\infty}^{\infty} \sum_{m=-\infty}^{\infty} c_{k-Nm} \ \delta(f - k) \\
  &amp;= \sum\limits_{k=-\infty}^{\infty} \tilde{c}_k \ \delta(f - k) \\
\end{align}$$

where

$$ \tilde{c}_k \triangleq \sum_{m=-\infty}^{\infty} c_{k-Nm} $$

so the coefficients of the Fourier series of the sampled periodic signal $x_\text{s}(t)$ are different from those of $x(t)$.  even in the baseband where the DTFT (and following, the DFT) sees the spectrum.