Function machine-2

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

$\begin{aligned} f(1-x) & = \frac {4^{1-x}}{4^{1-x} +2} & \small \blue{\text{Multiply up and down by }4^x} \\ & = \frac 4{4 +2(4^x)} & \small \blue{\text{Divide up and down by }2} \\ & = \frac 2{2 +4^x} \\ \implies f(x) + f(1-x) & = \frac {4^x}{4^x +2} + \frac 2{2 +4^x} = 1 \end{aligned}$

Then we have:

$\begin{aligned} f \left(\frac 1n\right) + f \left(\frac 2n\right) + \cdots + f \left(\frac {n-2}n\right) + f \left(\frac {n-1}n\right) & = 2019 \\ \implies \blue{f \left(\frac 1n\right)} + \red{f \left(\frac 2n\right)} + \cdots + \red{f \left(1-\frac 2n\right)} + \blue{f \left(1-\frac 1n\right)} & = 2019 \end{aligned}$

Since $f\left(\frac kn\right) + f\left(1-\frac kn\right) = 1$ , there are 2019 $f\left(\frac kn\right)$ - $f\left(\frac {n-k}n\right)$ pairs. That is 4038 $f(\cdot)$ terms or $n-1 = 4038 \implies n = \boxed{4039}$ .

We can rewrite $f(x) = 1 - \frac1{4^{x-1/2} + 1},$ so we have $\begin{aligned} \sum_{i=1}^{n-1} \left( 1 - \frac1{4^{i/n-1/2} + 1} \right) &= 2019 \\ n -1 - \sum_{i=1}^{n-1} \frac1{4^{i/n-1/2} + 1} &= 2019 \\ \sum_{i=1}^{n-1} \frac1{4^{i/n-1/2} + 1} &= n - 2020 \end{aligned}$ Let $a_i = \frac{i}{n} - \frac12$ be the exponent of $4$ in the above sum. Now the key observation is that $a_i = -a_{n-i},$ and $\frac1{4^{a_i}+1} + \frac1{4^{a_{n-i}} + 1} = \frac1{4^{a_i}+1} + \frac1{4^{-a_i} + 1} = 1.$ Using the substitution $i \mapsto n-i$ (which runs over the terms of the sum in reverse order), we can write $\begin{aligned} \sum_{i=1}^{n-1} \frac1{4^{a_i} + 1} &= n - 2020 \\ \sum_{i=1}^{n-1} \frac1{4^{a_{n-i}} + 1} &= n - 2020 \end{aligned}$ and adding these two equations gives $\begin{aligned} \sum_{i=1}^{n-1} 1 &= 2n-4040 \\ n-1 &= 2n-4040 \\ n &= \fbox{4039} \end{aligned}$