Evaluate DFT-downsampler-upsampler-IDFT block diagram
Given a finite length input sequence $x[n]$ of $N=64$ points, its 64-point DFT be $X[k]$. Then according to the block diagram you've provided, the signal $r[n]$ will be of $N/2$ points long and its $N/2$ - point DFT $R[k]$ will be related to $X[k]$ as :

$R[k] = X[2k]$, which will be manipulated as:

$$
\begin{align}
X[k] &amp;= \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N} kn} ~~~,~~~ k=0,1,...,N-1 \\
R[k] &amp; = X[2k] = \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N} 2kn} ~~~,~~~ k=0,1,...,N/2-1\\
&amp;=\sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N/2} kn} \\
&amp;=\sum_{n=0}^{\frac{N}{2}-1} x[n] e^{-j \frac{2\pi}{N/2} kn} + \sum_{n=\frac{N}{2}}^{N-1} x[n] e^{-j \frac{2\pi}{N/2} kn} \\
&amp;=\sum_{n=0}^{\frac{N}{2}-1} x[n] e^{-j \frac{2\pi}{N/2} kn} + \sum_{n=0}^{\frac{N}{2}-1} x[n+\frac{N}{2}] e^{-j \frac{2\pi}{N/2} kn} \\
&amp;=\sum_{n=0}^{\frac{N}{2}-1} \left( x[n]+x[n+\frac{N}{2}] \right) e^{-j \frac{2\pi}{N/2} kn} \\ 
&amp;= \text{N/2-point DFT} \{ r[n] \}\\ 
&amp;= \sum_{n=0}^{\frac{N}{2}-1} r[n] e^{-j \frac{2\pi}{N/2} kn} \\
\end{align}
$$  

hence we see that $r[n]$ is the sequence whose length (or period) is reduced to $N/2$ and is generated by adding the second half of $x[n]$ onto the first half.