IDFT of $Y[k]=2X[k]$ for even $k$
Below is a block-diagram of your signal processing chain (ignoring multiplication by 2).

$$ x[n] \longrightarrow \boxed{ \text{ DFT-N } } \longrightarrow X[k] \longrightarrow \boxed{ \downarrow 2 } \longrightarrow \boxed{ \uparrow 2 } \longrightarrow Y[k] \longrightarrow \boxed{ \text{ IDFT-N } } \longrightarrow y[n]$$

Where sequence $x[n]$ has a length of $N=16$ samples, $X[k]$ is its $N$-point DFT sequence. And $Y[k]$ is related to $X[k]$ as given by 
$$ Y[k] = \begin{cases} { 2 X[k] ~~~ , ~~~ k = 0,2,...,N-2 \\ 0 ~~~ , ~~~ \text{ otherwise } \\} \end{cases} $$

I will solve this problem using the duality argument. Eventhough it's not necessary, it can reduce the mathematical burden via bypassing explicit computations.

The duality in DFT stems from the fact that the forward and inverse DFT transforms are (almost) identical, designated with the following:  ($n$ time , $k$ frequency) 

$$ x_t[n] \leftrightarrow X_f[k] \iff X_f[n] \leftrightarrow N x_t[-k] $$  

We can use the duality property to observe that if an operation  $T$ is performed on the time-domain signals $x[n] \xrightarrow{T} y[n]$ with a corresponding reflection on the frequency domain as $X[k] \xrightarrow{M} Y[k]$, where $M$ is the frequency-domain reflection operator of the time-domain operator $T$ , then the same operation $T$ performed on the freqency domain signals will have the same reflection operator $M$ on the corresponding time domain signals, with a possible scale factor if different sample rates are used on the left and right nodes. 

Therefore, to use duality, in the above block diagram, I'll replace the frequency domain relation between $Y[k]$ and $X[k]$ with a time domain relation as: 

$$v[n] \longrightarrow \boxed{ \downarrow 2 } \longrightarrow w[n] \longrightarrow \boxed{ \uparrow 2 } \longrightarrow z[n] $$

where in the time domain the operation is compression (downsampling) and expansion (upsampling) whose frequency domain DFT relation is readily available as: 

$$ W[k] = 0.5 ( V[k] + V[k+N/2] ) ~~~, k = 0,1,..,N/2-1 $$

where $W[k]$ is $N/2$ point DFT of $N/2$ point compressed sequence $w[n]$, and $V[k]$ is $N$ point DFT of $N$ point sequence $v[n]$. Stated verbally, $W[k]$ is obtained from $V[n]$ by adding its second half onto its first half.

Then 
$$Z[k] = W[ (k)_{N/2}] ~~~, k = 0,1,2,...,N-1 $$

Where $Z[k]$ is the $N$ point DFT of length $N$ point sequence $z[n]$. Again stated verbally, $Z[k]$ is merely a repetation of $W[k]$ two times. This is signified by the ramge of $k=0,1,2,...,N-1$ and the modulus operator inside the argument of $W[k]$.

Hence we can find the relation between $Z[k]$ and $V[k]$ as :
$$Z[k] = 0.5 \left( V[ (k)_{N/2}] + V[ (k)_{N/2} + N/2] \right) $$ 

This relation in the frequency domain was readily available. I didn't use any derivations to achieve it (eventhough at first it was derived by means of DFT manipulations anyway). Now using the duality property of DFT, I assert that, if te time domain signals are replaced by frequency domain signals then their effect on the time domain will similar to the above frequency domain relation, by simply replacing $k$ with $n$. Note that time reversal not needed, as it will be cancelled, but scaling by $2$ is required. So the result will be: 

$$ \boxed{ y[n] =   x[ (n)_{N/2}] + x[ (n)_{N/2} + N/2] } $$

where $N=16$, $n=0,1,2,...,15$.

Conclusion. It takes more effort to describe a solution more easily obtained via the duality principle.