Why the following discrete type system,variable accumulator, is time invariant?
You are overloading the symbol $n$ which doesn't mean the same everywhere in your calculations.  Instead, write down what the sequence of input values is and the corresponding sequence of output values when the input is $x$.  DO NOT use the symbol $n$ anywhere ($n_0$ is OK to use).  So, your answer should look like

\begin{align}
y[-2] &amp;= \cdots\\
y[-1] &amp;= \cdots\\
y[0] &amp;= x[-n_0] + x[-n_0+1] + \cdots + x[0] + x[1] + \cdots x[n_0]\\
y[1] &amp;= x[-n_0+1] + \cdots + x[0] + x[1] + \cdots x[n_0+1]
\end{align}
etc.

Now, shift $x$ by $k$ places (say $k=1$ for starters) and call the result $\hat x$. Repeat the above computation for the input $\hat x$ and call the result $\hat y$.  Is $\hat y$ the same as $y$ except shifted by $k$ places? Repeat for $k+1$ etc until you have managed to convince yourself that regardless of what $k$ you choose, shifting $x$ by $k$ places results in $y$ also shifting by $k$ places.