Why does adding a time delayed version of a signal to itself create a filtered signal?
When you delay a signal by $T$ seconds and add it to the signal itself,
you are cancelling out or nulling the signal component at frequency 
$\frac{1}{2T}$ Hz since that signal component will have changed phase
by exactly $\pi$:
$$\begin{align}
\sin\left(2\pi\frac{1}{2T}t + \theta\right) +
\sin\left(2\pi\frac{1}{2T}(t-T) + \theta\right)
&amp;= \sin\left(2\pi\frac{1}{2T}t + \theta\right)
+ \sin\left(2\pi\frac{1}{2T}t + \theta  - \pi\right)\\
&amp;= \sin\left(2\pi\frac{1}{2T}t + \theta\right)
+\sin\left(2\pi\frac{1}{2T}t + \theta\right)\cos(\pi)\\
&amp;\ \hspace{0.2in}-\cos\left(2\pi\frac{1}{2T}t + \theta\right)\sin(\pi)\\
&amp;= \sin\left(2\pi\frac{1}{2T}t + \theta\right)
-\sin\left(2\pi\frac{1}{2T}t + \theta\right)-0\\
&amp;= 0.
\end{align}$$
A similar thing happens at odd multiples of $\frac{1}{2T}$ Hz also.
For nearby frequencies, the cancellation is not as complete, and
of course, at even multiples of $\frac{1}{2T}$ Hz, the signal 
component is doubled in value instead of being cancelled.
Similarly, if the delayed signal is reduced in amplitude,
cancellation is not complete at $\frac{1}{2T}$ Hz etc.

To summarize, the signal is being filtered because different
frequencies are being passed through with different gains.

If you want the frequency-domain explanation, the transfer
function $H(f)$ of the system is the Fourier transform of what Matt's
answer gave as the impulse response, viz.
$$\mathcal F\left[\delta(t) + \delta(t-T)\right] = 1+\exp(-j2\pi fT)$$
which is a nonconstant function of $f$ (in fact, $|H(f)|$ varies
sinusoidally from a maximum of $2$ to a minimum of $0$ as discussed
above), and so $Y(f)=H(f)X(f)$ is
not a scalar multiple of $X(f)$. Filtering!