Let $f:R\to R$ be a function defined $f\left(x\right)=\frac{2e^{2x}}{e^{2x}+e}$. Then $f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+f\left(\frac{3}{100}\right)+......\dots ..+f\left(\frac{99}{100}\right)$ is equal to ________

$\begin{array}{l}f\left(x\right)+f(1-x)=\frac{2e^{2x}}{e^{2x}+e}+\frac{2e^{2-2x}}{e^{2-ex}+e}=\left[\frac{e^{2x}}{e^{2x}+e}+\frac{e^2}{e^2+e^{2x+1}}\right]\\ \\ =2\left[\frac{e^{2x-1}}{e^{2x-1}+1}+\frac{1}{1+e^{2x-1}}\right]=2\\ \\ f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+f\left(\frac{3}{100}\right)+\dots .....+f\left(\frac{99}{100}\right)\\ \\ =\left\{f\left(\frac{1}{100}\right)+f\left(\frac{99}{100}\right)\right\}+\left\{f\left(\frac{2}{100}\right)+f\left(\frac{98}{100}\right)\right\}+\dots .....+f\left\{\left(\frac{49}{100}\right)+f\left(\frac{51}{100}\right)\right\}+f\left(\frac{1}{2}\right)\\ \\ =(2+2+2+----49 times)+\frac{2e}{e+e}\\ \\ =98+1=99\end{array}$