We change the amphitheater into a $21\times 21$ metrices with the last column and row are colored red \[\begin{bmatrix} *&*&*&*&*&\textcolor{red}{*}\\ *&*&*&*&*&\textcolor{red}{*}\\ *&*&*&*&*&\textcolor{red}{*}\\ *&*&*&*&*&\textcolor{red}{*}\\ *&*&*&*&*&\textcolor{red}{*}\\ \textcolor{red}{*}&\textcolor{red}{*}&\textcolor{red}{*}&\textcolor{red}{*}&\textcolor{red}{*}&\textcolor{red}{*} \end{bmatrix}_{21\times21}\]The value $a_{i,j}=1$ and $0$ if the person in that seat raises blue and yellow sign respectively. We claim that No matter how the $20\times 20$ submetrices with black $*$ are labelled with $0,1$, there is exactly one way to label the rest $41$ red $*$. which is true since the parity of the sum of each row and column must be odd. Thus, there are $2^{20\times 20}=\boxed{2^{400}}$ ways to raise signs.