Is the sum of two white noise processes necessarily a white noise? Let $a_t$ and $b_t$ be white noise processes. Can we say $c_t=a_t+b_t$ is necessarily a white noise process?
 A: No, you need more (at least under Hayashi's definition of white noise). For example, the sum of two independent white noise processes is white noise.
Why is $a_t$ and $b_t$ white noise insufficient for $a_t+b_t$ to be white noise?
Following Hayashi's Econometrics, a covariance stationary process $\{z_t\}$ is defined to be white noise if $\mathrm{E}[z_t] = 0$ and $\mathrm{Cov}\left(z_t, z_{t-j} \right) = 0$ for $j \neq 0$.
Let $\{a_t\}$ and $\{b_t\}$ be white noise processes. Define $c_t = a_t + b_t$. Trivially we have $\mathrm{E}[c_t] = 0$. Checking the covariance condition:
\begin{align*}
\mathrm{Cov} \left( c_t, c_{t-j} \right) &= \mathrm{Cov} \left( a_t, a_{t-j}\right) + \mathrm{Cov} \left( a_t, b_{t-j}\right) + \mathrm{Cov} \left( b_t, a_{t-j}\right) + \mathrm{Cov} \left( b_t, b_{t-j}\right)
\end{align*}
Applying that $\{a_t\}$ and $\{b_t\}$ are white noise:
\begin{align*}
\mathrm{Cov} \left( c_t, c_{t-j} \right) &=  \mathrm{Cov} \left( a_t, b_{t-j}\right) + \mathrm{Cov} \left( b_t, a_{t-j}\right) 
\end{align*}
So whether $\{c_t\}$ is white noise depends on whether $\mathrm{Cov} \left( a_t, b_{t-j}\right) + \mathrm{Cov} \left( b_t, a_{t-j}\right) = 0$ for all $j\neq 0$.
Example where sum of two white noise processes is not white noise:
Let $\{a_t\}$ be white noise. Let $b_t = a_{t-1}$. Observe that process $\{b_t\}$ is also white noise. Let $c_t = a_t + b_t$, hence  $c_t = a_t + a_{t-1}$, and observe that process $\{c_t\}$ is not white noise.
A: Even simpler than @MatthewGunn's answer,
Consider $b_t = -a_t$. Obviously $c_t \equiv 0$ is not white noise -- it'd be hard to call it any kind of noise.
The broader point is, if we don't know anything about the joint distribution of $a_t$ and $b_t$, we won't be able to say what happens when we try and examine objects which depend on both of them. The covariance structure is essential to this end.

Addendum:
Of course, this is exactly the purpose of noise-cancelling headphones! -- to reverse the frequency of external noises and cancel them out -- so, going back to the physical definition of white noise, this sequence is literal silence. No noise at all.
A: In electronics, white noise is defined as having a flat frequency spectrum ('white') and being random ('noise'). Noise generally can be contrasted with 'interference', one or more undesired signals being picked up from elsewhere and being added to the signal of interest, and 'distortion', undesired signals being generated from nonlinear processes acting on the signal of interest itself. 
While it is possible for two different signals to have correlated parts, and therefore cancel differently at different frequencies or at different times, e.g. completely canceling over a certain band of frequencies or during a certain interval of time, but then not canceling, or even adding constructively over another band of frequencies or during a certain interval of time, the correlation between the two signals presumes a correlation, which is precluded by the presumably random aspect of 'noise', which is what was asked about.  
If, indeed, the signals are 'noise' and therefore independent and random, then no such correlations should/would exist, so adding them together will also have a flat frequency spectrum and will therefore also be white.  
Also, trivially, if the noises are exactly anti-correlated, then they could cancel to give zero output at all times, which also has a flat frequency spectrum, zero power at all frequencies, which could fall under a sort of degenerate definition of white noise, except that it isn't random and can be perfectly predicted. 
Noise in electronics can come from several places.  For example, shot noise, arising from the random arrival of electrons in a photocurrent (coming from the random arrival times of photons), and Johnson noise, coming from the Brownian motion of electrons in a resistive element warmer than absolute zero, both produce white noise, although, always with a finite bandwidth at both ends of the spectrum in any real system measured over a finite length of time.
A: if both white noise sound is traveling in same direction And if their frequency is in phase matched up, then only they get added. But, one thing i am not sure about is after adding up will it remain as white noise or it will become some other type of sound having different frequency.
