ouput of low pass filter
The half-band lowpass filter, has the following impulse response :
$$ h[n] = \frac{\sin(\frac{\pi}{2} n) }{ \pi n} $$

And for $x[n] = \delta[n]$ ,  $w[n] = h[n]$.

Then the output $y[n]$ will be :

$$y[n] = w[n] \big( 1 + (-1)^n \big) $$

which can be simplified by expanding the parenthesis as:

$$ 
\begin{align}
y[n] &amp;= \frac{\sin(\frac{\pi}{2} n) }{ \pi n} \cdot \big( e^{j 2 \pi n} + e^{j \pi n} \big) \\ \\
y[n] &amp;= \frac{\sin(\frac{\pi}{2} n) }{ \pi n} \cdot  e^{j \frac{3 \pi}{2} n} \cdot  \big( e^{j \frac{\pi}{2} n} + e^{-j \frac{\pi}{2} n} \big)\\ \\
y[n] &amp;= \frac{ \sin(\frac{\pi}{2} n) }{ \pi n} \cdot 2 \cdot  \cos( \frac{\pi}{2} n) \cdot  e^{j \frac{3 \pi}{2}  n} \\ \\
y[n] &amp;= \frac{\sin(\pi n) }{ \pi n}  \cdot  e^{j \frac{3 \pi}{2} n}
\end{align} 
$$
Which can be shown to be :

$$y[n] =  \begin{cases}{ 1 ~~~~, n = 0 \\ 0 ~~~~ , n \neq 0  } \end{cases} $$

Therefore $y[n] = \delta[n]$. The system is a zero-phase all pass filter. And finally, the required quantity is 
$$ M = \sum_{n=-\infty}^{\infty} y[n] = \sum_{n=-\infty}^{\infty} \delta[n] = 1 $$