Alternating sign after Discrete Fourier transform
Let's suppose that your original data, $u_n$, has been windowed (which is something you might typically do, in order to 'tidy up' the FFT). In this case, most of the energy in your input data sequence is clustered around the region near $n=N/2$. To be more specific, I expect that your input data is actually centred on $n=N/2+1$.

Now, look at your expression for the FFT, and imagine that all the $u_n$ are zero, except $u_{N/2+1}=1$. This means that:

$v_s=\frac{1}{N}\sum_{r=1}^N u_r \exp[2\pi i(r−1)(s−1)/n] = \frac{1}{N} \exp[2\pi i(N/2)(s-1)] = \frac{1}{N} (-1)^{s-1}$

In other words, the sign-flipping behaviour of your FFT output is generally related to the fact that the energy of you input sequence, $u_n$, is clustered around time $N/2$.

Ideally, you'd like to have your energy clustered around time zero, and in order to do this, you could shift your signal $N/2$ samples to the left (this should be a circular shift, so that the data wraps around in  the buffer). In practice, you don't bother with the circular shift, and instead you simply swap the sign of all the odd FFT bins, which does exactly the same thing as shifting the data in the time domain.

Also, if your FFT output is real, this implies that your time-domain input is Hermitian. This means that there is a particular symmetry in your input data, $u_n$, such that the real part of the input satisfies:

$Re(u_{N/2+1+k}) = Re(u_{N/2+1-k})$ for all $k$ $\quad\leftarrow$ This means that the real part is even

and the odd part satisfies:

$Im(u_{N/2+1+k}) = -Im(u_{N/2+1-k})$ for all $k$ $\quad\leftarrow$ This means that the imaginary part is odd

If you look at your data (the $u_n$ values) you should see this property. You will find that $u_{N/2+1-2}$ (which is real, according to your description) will be equal to $u_{N/2+1+2}$. Furthermore, $u_{N/2+1-1}$ (which is imaginary, according to your description) will be equal to $-u_{N/2+1+1}$.