f(x) = \(\frac{e^{2x-1}}{1+e^{2x-1}}\)
f(1-x) = \(\frac{e^{2(1-x)-1}}{1+e^{2(1-x)-1}}\) = \(\frac{e^{1-2x}}{1+e^{1-2x}}\) = \(\frac{e^{-(2x-1)}}{1+e^{-(2x+1)}}\)
= \(\frac1{e^{2x-1}+1}=\frac1{1+e^{2x-1}}\)
\(\therefore\) f(x) + f(1 - x) = \(\frac{e^{2x-1}+1}{1+e^{2x-1}}=1\)
\(\therefore\) f(1/2009) + f(2008/2009) = 1
\(\therefore\) f(2/2009) + f(2007/2009) = 1
f(1004/2009) + f(1005/2009) = 1
\(\therefore\) f(1/2009) + f(2008/2009) + f(2/2009) + f(2007/2009) +....f(1004/2009) +f(1005/2009)
= 1 + 1 + ...1 (10004 times) = 1004
⇒ f(1/2009) + f(2/2009) + .... + f(1004/2009) + f(1005/2009) + ....+ f(2007/2009) + f(2008/2009) = 1004