jmc

The graphs of $y=f(x)$ and $y=f^{[2]}(x)$ are shown below.



For $n\ge 2,$ $$f^{[n]}(x)=f^{[n-1]}(f(x))=\{\begin{array}{cl}{f}^{[n-1]}(2x)& \text{if 0≤x≤21},\\ f^{[n-1]}(2-2x)& \text{if 21≤x≤1}.\end{array}$$Let $g(n)$ be the number of values of $x\in [0,1]$ for which $f^{[n]}(x)=\frac{1}{2}.$ Then $f^{[n]}(x)=\frac{1}{2}$ for $g(n-1)$ values of $x\in \left[0,\frac{1}{2}\right],$ and $g(n-1)$ values of $x$ in $\left[\frac{1}{2},1\right].$

Furthermore $$f^{[n]}\left(\frac{1}{2}\right)=f^{[n]}(1)=0\ne \frac{1}{2}$$for $n\ge 2.$ Hence, $g(n)=2g(n-1)$ for all $n\ge 2.$ Since $g(1)=2,$ $g(2005)=2^{2005}.$ The final answer is $2+2005=\boxed{2007}.$