1. We need to find $S= f\left(\frac1{20}\right)+f\left(\frac2{20}\right)+\cdots+f\left(\frac{39}{20}\right),$ where $f(x)=\frac{5^x}{5^x+5}.$

2. First, simplify the function in a useful way.

Divide numerator and denominator by $5^x$: $f(x)=\frac{1}{1+5^{1-x}}.$

Now compute $f(1-x)$: $f(1-x)=\frac{5^{1-x}}{5^{1-x}+5}.$ Divide numerator and denominator by $5$: f(1-x)=\frac{5^{-x}}{5^{-x}+1}=rac{1}{1+5^x}.

Also, f(x)=\frac{5^x}{5^x+5}=rac{5^{x-1}}{5^{x-1}+1}=rac{1}{1+5^{1-x}}.

So, $f(x)+f(1-x)=\frac{1}{1+5^{1-x}}+\frac{1}{1+5^x}.$ But an easier direct check is: Let $t=5^x$. Then f(x)=\frac{t}{t+5},\qquad f(1-x)=\frac{5/t}{5/t+5}=rac{1}{t+1}. Hence $f(x)+f(1-x)=\frac{t}{t+5}+\frac{1}{t+1}.$ This is not immediately $1$, so let us instead rewrite correctly:

Since $f(x)=\frac{5^x}{5^x+5}=\frac{5^{x-1}}{5^{x-1}+1},$ put $u=5^{x-1}$. Then $f(x)=\frac{u}{u+1}.$ Now $f(2-x)=\frac{5^{2-x}}{5^{2-x}+5}=\frac{5^{1-x}}{5^{1-x}+1}.$ But since $u=5^{x-1}$, we have $5^{1-x}=\frac1u$. Therefore $f(2-x)=\frac{1/u}{1/u+1}=\frac{1}{u+1}.$ Thus, $f(x)+f(2-x)=\frac{u}{u+1}+\frac{1}{u+1}=1.$

So the key identity is: $\boxed{f(x)+f(2-x)=1}.$

3. Now look at the terms of the series: $\frac1{20},\frac2{20},\frac3{20},\dots,\frac{39}{20}.$ There are $39$ terms.

Pair them as: $f\left(\frac1{20}\right)+f\left(2-\frac1{20}\right)=f\left(\frac1{20}\right)+f\left(\frac{39}{20}\right)=1,$ $f\left(\frac2{20}\right)+f\left(\frac{38}{20}\right)=1,$ $f\left(\frac3{20}\right)+f\left(\frac{37}{20}\right)=1,$ and so on.

Thus the terms pair up to give $1$ each.

4. Since there are $39$ terms, one middle term remains unpaired: $\frac{20}{20}=1.$ So, $f(1)=\frac{5^1}{5^1+5}=\frac{5}{10}=\frac12.$

The remaining $38$ terms make $19$ pairs, each summing to $1$. Therefore, $S=19+\frac12=\frac{39}{2}.$

5. Hence the correct option is $\boxed{\frac{39}{2}}.$