If $\frac{1}{(20-\mathrm{a})(40-\mathrm{a})}+\frac{1}{(40-\mathrm{a})(60-\mathrm{a})}+\dots +\frac{1}{(180-\mathrm{a})(200-\mathrm{a})}=\frac{1}{256}$then the maximum value of a is:

$\frac{1}{(20-\mathrm{a})(40-\mathrm{a})}+\frac{1}{(40-\mathrm{a})(60-\mathrm{a})}+\dots .+\frac{1}{(180-\mathrm{a})(200-\mathrm{a})}=\frac{1}{256}$
$\Rightarrow \frac{1}{20}\left[\frac{(40-\mathrm{a})-(20-\mathrm{a})}{(20-\mathrm{a})(40-\mathrm{a})}+\frac{(60-\mathrm{a})-(40-\mathrm{a})}{(40-\mathrm{a})(60-\mathrm{a})}\right.\left.+\dots +\frac{(200-\mathrm{a})-(180-\mathrm{a})}{(180-\mathrm{a})(200-\mathrm{a})}\right]=\frac{1}{256}$
$\Rightarrow \frac{1}{20}\left[\frac{1}{(20-\mathrm{a})}-\frac{1}{(40-\mathrm{a})}+\frac{1}{(40-\mathrm{a})}-\frac{1}{(60-\mathrm{a})}\right.\left.+\dots +\frac{1}{(180-\mathrm{a})}-\frac{1}{(200-\mathrm{a})}\right]=\frac{1}{256}$
$\begin{array}{l}\Rightarrow \frac{1}{20}\left[\frac{1}{20-\mathrm{a}}-\frac{1}{200-\mathrm{a}}\right]=\frac{1}{256}\\ \Rightarrow \frac{1}{20}\left[\frac{(200-\mathrm{a})-(20-\mathrm{a})}{(20-\mathrm{a})(200-\mathrm{a})}\right]=\frac{1}{256}\\ \Rightarrow \frac{180}{20(20-\mathrm{a})(200-\mathrm{a})}=\frac{1}{256}\end{array}$
$\begin{array}{l}\Rightarrow (20-\mathrm{a})(200-\mathrm{a})=9\times 256\Rightarrow 4000-220\mathrm{a}+{\mathrm{a}}^2=2304\\ \Rightarrow {\mathrm{a}}^2-220\mathrm{a}+1696=0\Rightarrow {\mathrm{a}}^2-212\mathrm{a}-8\mathrm{a}+1696=0\\ \Rightarrow (\mathrm{a}-212)(\mathrm{a}-8)=0\Rightarrow \mathrm{a}=8,212\end{array}$
$\mathrm{Hence}, \mathrm{maximum} \mathrm{value} \mathrm{of} \mathrm{a} \mathrm{is} 212.$