DFT sample point k < N has negative frequency
The DTFT frequency $\omega_k$ corresponding to the $N$-point DFT index $k$ is given by (frequency sampling relation)

$$ \omega_k = \frac{2 \pi}{N} k ~~~,~~~ k = 0,1,...,N-1.$$

The confusion might arise with the interpretation that the, on the continuous-time domain these frequencies correspond to:

$$  f_k = \frac{f_s}{N} k ~~~,~~~ k = 0,1,...,N-1 ?$$

But no. Because, after about $k = N/2$ which corresponds to $\omega = \pi$ and $f = f_s/2$, you will be sampling the negative portion of the shifted spectrum: $X(e^{j (\omega-2\pi)})$, which corresponds to negative discrete-time frequencies of $X(e^{j \omega})$ due to the periodicity of DTFT $X(e^{j\omega})$.

A more relevant (eventually the same) interpretation of the CTFT, DTFT, DFT frequency relation can be given by the following. 

Let $x_c(t)$ be bandlimited to $-B \leq f &lt;B$ and sampled at $F_s = 2B$ samples per second. The base period of the spectrum of $x[n]$ is between $-\pi \leq \omega &lt; \pi$. Note that only the left member is included and the right member is discluded. 

Now assume an $N$-point DFT $X[k]$ of $x[n]$ is computed. According to the frequency sampling interpretation of DFT, samples of $X[k]$ are given by :

$$ X[k] = X(e^{j \frac{2\pi}{N} k}) ~~~~ , k = ...-3,-2,-1,0,1,2,3... $$

Note very carefully, that eventhough most of the time we restrict our attention of the DFT index $k$ into the range of $0 \leq k &lt; N$, indeed, the relation is valid for all $k$ which is a manifestation of DFT $X[k]$ and DTFT $X(e^{j \omega})$ being periodic by $N$ and $2\pi$. 

Hence, instead of taking the range of $k$ in $[0,N-1]$, we can also take it in the range: $[-N/2 , N/2-1]$ (assuming $N$ even for the moment). 

So, if you take $k = -N/2,-N/2+1,...,-1,0,1,2,...,N/2-1$ as the range of $k$ for DFT $X[k]$, then the mapping into the negative frequencies become obvious.

Note that, conventional FFT software always take the range of $k$ in $[0,N-1]$. Therefore, one has to relate them too which is simple: for $k$ in $[-N/2,-1]$ it corresponds to $k'$ in $[N/2,N-1]$ in the conventional DFT range, and for $k$ in $[0,N/2-1]$ it corresponds to the $k'$ in $[0,N/2-1]$ range in the conventional range.