Happy New Year function😆

$given$

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

$f(1-x) = \frac{4^{1-x}}{4^{1-x} +2}$

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space = \frac{\frac{4}{4^x}}{\frac{4}{4^x}+2}$

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space = \frac{\frac{4}{4^x}}{\frac{4+2.4^x}{4^x}}$

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space = \frac{4}{4^x} \times \frac{4^x}{4+2.4^x}$

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space = \frac{4}{4+2.4^x}$

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space = \boxed{ \frac{2} {4^x+2}}$

$Now$

$f(x)+f(1-x)$ $=\frac{4^x+2}{4^x+2} =\boxed{ 1}$

$1st + last = f(\frac{1}{2019})+f(\frac{2018}{2019})$

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space= f(\frac{1}{2019})+f(1-\frac{1}{2019})$

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space= 1$ $\space \space \space \space \space \space \space \space$ $\boxed{ \therefore f(x)+f(1-x)=1 }$

$we \space can \space see \space that, \space sum \space of \space every \space pair \space from \space given \space series \space is \space \boxed{ 1 }$

$So,$

$Total \space Pair \space = \space 2018÷2$

$\space\space\space\space\space\space\space\space \space \space \space \space \space \space \space \space \space \space \space \space = \boxed{ 1009}$

So, The final answer is $\boxed {1009}$

Given that $f(x) = \dfrac {4^x}{4^x+2}$ , then

$\begin{aligned} f(x) + f(1-x) & = \frac {4^x}{4^x+2} + \color{#3D99F6} \frac {4^{1-x}}{4^{1-x}+2} & \small \color{#3D99F6} \text{Multiply up and down by }4^{x-\frac 12}. \\ & = \frac {4^x}{4^x+2} + \color{#3D99F6} \frac {2}{2+4^x} \\ & = 1 \end{aligned}$

Now we have:

$\begin{aligned} S & = f \left(\frac 1{2019}\right) + {\color{#3D99F6} f \left(\frac 2{2019}\right)} + {\color{#D61F06} f \left(\frac 3{2019}\right)} + \cdots + {\color{#D61F06} f \left(\frac {2016}{2019}\right)} + {\color{#3D99F6} f \left(\frac {2017}{2019}\right)} + f \left(\frac {2018}{2019}\right) \\ & = f \left(\frac 1{2019}\right) + {\color{#3D99F6} f \left(\frac 2{2019}\right)} + {\color{#D61F06} f \left(\frac 3{2019}\right)} + \cdots + {\color{#D61F06} f \left(1- \frac 3{2019}\right)} + {\color{#3D99F6} f \left(1- \frac 2{2019}\right)} + f \left(1 - \frac 1{2019}\right) \\ & = \underbrace{1 + 1 + 1 + \cdots + 1}_{\text{Number of 1s}=1009} \\ & = \boxed{1009} \end{aligned}$

This is not a real solution, it's just a way to solve the problem looking to options. See Jahangir and Otto's comments for effective solutions.

For all $x$ it's true that

$\frac{4^x}{4^x+2}<1$

so must hold also

$\sum_{i=1}^{2018}f(\frac{i}{2019})<2018$ .

Now, we have only two options for the solution, $0$ and $1009$ . But all terms are positive, so only one option is possible: $\boxed{1009}$ .

Just to have some New Year's fun, we can write $f(x)=\frac{1}{2}\tanh\left(\ln 2(x-\frac{1}{2})\right)+\frac{1}{2}$ . Since $\tanh t$ is an odd function and the given points are arranged symmetrically with respect to $x=\frac{1}{2}$ , the answer is $2018\times \frac{1}{2}=\boxed{1009}$ .