Given,

$$f(x) = {{2{e^{2x}}} \over {{e^{2x}} + e}}$$

$$\therefore$$ $$f(1 - x) = {{2{e^{2(1 - x)}}} \over {{e^{2(1 - x)}} + e}}$$

$$ = {{2\,.\,{{{e^2}} \over {{e^{2x}}}}} \over {{{{e^2}} \over {{e^{2x}}}} + e}}$$

$$ = {{2{e^2}} \over {{e^2} + {e^{2x}}\,.\,e}}$$

$$ = {{2{e^2}} \over {e(e + {e^{2x}})}}$$

$$ = {{2e} \over {e + {e^{2x}}}}$$

$$\therefore$$ $$f(x) + f(1 - x) = {{2{e^{2x}}} \over {{e^{2x}} + e}} + {{2e} \over {{e^{2x}} + e}}$$

$$ = {{2({e^{2x}} + e)} \over {{e^{2x}} + e}}$$

$$ = 2$$ ...... (1)

Now,

$$f\left( {{1 \over {100}}} \right) + f\left( {{{99} \over {100}}} \right)$$

$$ = f\left( {{1 \over {100}}} \right) + f\left( {1 - {1 \over {100}}} \right)$$

$$ = 2$$ [as $$f(x) + f(1 - x) = 2$$]

Similarly,

$$f\left( {{2 \over {100}}} \right) + f\left( {1 - {2 \over {100}}} \right) = 2$$

$$ \vdots $$

$$f\left( {{{49} \over {100}}} \right) + f\left( {1 - {{49} \over {100}}} \right) = 2$$

$$\therefore$$ Total sum $$ = 49 \times 2$$

Remaining term $$ = f\left( {{{50} \over {100}}} \right) = f\left( {{1 \over 2}} \right)$$

Put $$x = {1 \over 2}$$ in equation (1), we get

$$f\left( {{1 \over 2}} \right) + f\left( {1 - {1 \over 2}} \right) = 2$$

$$ \Rightarrow 2f\left( {{1 \over 2}} \right) = 2$$

$$ \Rightarrow f\left( {{1 \over 2}} \right) = 1$$

$$\therefore$$ Sum $$ = 49 \times 2 + 1 = 99$$