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We have, $f(x)=\frac{4^x}{4^x+2}$

Now, $f(1-x)=\frac{4^{1-x}}{4^{1-x}+2}=\frac{4}{4+2\cdot4^x}$ $\Rightarrow f(1-x)=\frac{2}{2+4^x}$ Now we observe a very nice property of the function from above which is, $f(x)+f(1-x)= 1\text(why)$ Now a sudden consequence of this implies, $\Rightarrow f(\frac{1}{999})+f(\frac{998}{999})=1, f(\frac{2}{999})+f(\frac{997}{999})=1$ $--- f(\frac{499}{999})+f(\frac{500}{999})=1$ $\Rightarrow \sum_{k=1}^{998} f(\frac{k}{999})=1+1+++(499 terms)=\boxed{499}$ Note that there are a class of functions that satisfy this property. The general form of such functions for integer a is: $\huge\boxed{f(x)=\frac{a^x}{a^x+\sqrt{a}}}$