Upsampling DFT at frequency domain
It can mean the following: 
Consider any signal $x_{N}[n]$ of length $N$, and form the length $2N$ signal $x_{2N}[n]$ by repeating a copy of $x_{N}[n]$, then compute $2N$-Point DFT of this new signal $x_{2N}[n]$ whose every odd bin will be zero. i.e. $$X_{2N}[k] = 0 ~~,~~ \text{for}~~ k=1,3,5...,2N-1$$  

whose demonstration can be performed with the following Octave/Matlab code:

 N = 16;
 x = randn(1,N);         % form an arbitrary signal x[n]
 x2 = [x x];             % repeat it once
 X2 = fft(x2,2*N);       % take DFT of length 2N
 figure,stem(abs(X2));   % display the result


Therefore it can mean that if you are performing an $M$ point DFT on your original signal $x[n]$, the signal $x[n]$ could actually be periodic with $\frac M2$ for $M$ even.

A more mathematical derivation of the result can be seen as follows:

Consider a signal $x_N[n]$ of length $N$ and its repetition $x_{2N}[n]$ of length $2N$ such that $x_{2N}[n]=x_N[n \mod N]$ for $n=0,1,\ldots,2N-1$.

Let's denote $N$-Point DFT of $x_N[n]$ as $X_N[k]$ for $k=0,1,\ldots,N-1$ and $2N$-Point DFT of $x_{2N}[n]$ as $X_{2N}[k]$ for $k=0,1,\ldots,2N-1$:  

\begin{align}
&amp;{\text{Start with defining $2N$-Point DFT of $x_{2N}[n]\ldots$}}\\
X_{2N}[k] &amp;= \sum_{n=0}^{2N-1} {x_{2N}[n] e^{-j\frac{2\pi}{2N}nk}} \\ \\
&amp;{\text{first split the sum and then substitude $n=n-N$ in the $2^{nd}$}}\\
&amp;= \sum_{n=0}^{N-1} {x_{2N}[n] e^{-j\frac{2\pi}{N}n(k/2)}} + \sum_{n=N}^{2N-1} {x_{2N}[n] e^{-j\frac{2\pi}{2N}nk}} \\ \\ 
&amp;{\text{recognise $x_{2N}[n]=x_N[n]$, $x_{2N}[n+N]=x_N[n]$ }}\\
&amp;= \sum_{n=0}^{N-1} {x_{2N}[n] e^{-j\frac{2\pi}{N}n(k/2)}} + \sum_{n=0}^{N-1} {x_{2N}[n+N] e^{-j\frac{2\pi}{2N}(n+N)k}} \\ \\ 
&amp;{\text{Expand the 2nd sum's multiplier factor}}\\ 
&amp;= \sum_{n=0}^{N-1} {x_N[n] e^{-j\frac{2\pi}{N}n(k/2)}} + \sum_{n=0}^{N-1} {x_N[n] e^{-j\frac{2\pi}{N}(n+N)(k/2)}} \\ \\ 
&amp;{\text{Replace $e^{-2j\frac\pi N}$ with $W_N$ for simplicity }} \\
&amp;= \sum_{n=0}^{N-1} {x_N[n] W_N^{nk/2}} + W_N^{Nk/2}\sum_{n=0}^{N-1} {x_N[n] W_N^{nk/2}} \\ \\ 
&amp;{\text{Recognise the sums as $N$-Point DFT $X[k]$ of $x[n]$, at $\frac k2$}}\\
&amp;= X_N[k/2] + e^{-j\pi k} X_N[k/2] \\ \\ 
&amp;= X_N[k/2] \cdot \big( 1 + (-1)^k \big)  \\ \\
&amp;= \begin{cases}
2 X_N[k/2] , &amp;\scriptstyle{\text{k=0,2,4,...,2N-2 }}\\
0  , &amp;\scriptstyle{\text{k=1,3,5,...,2N-1 }}\\
\end{cases}
\end{align}

Shows that for such a defined signal , its odd DFT bins will be zero...