Given,

$f\left(x\right)=\frac{2^{2x}}{2^{2x}+2}$,

so $f\left(1-x\right)=\frac{2^{2-2x}}{2^{2-2x}+2}=\frac{2^2}{2^2+2^{2x+1}}=\frac{2}{2+2^{2x}}$

Now $f\left(x\right)+f\left(1-x\right)=\frac{2^{2x}}{2^{2x}+2}+\frac{2}{2+2^{2x}}=1$

So by using the relation $f\left(x\right)+f\left(1-x\right)=1$ we will solve the given equation 

$f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+f\left(\frac{3}{2023}\right).........f\left(\frac{2022}{2023}\right)$

$=f\left(\frac{1}{2023}\right)+f\left(\frac{2022}{2023}\right)+f\left(\frac{2}{2023}\right)+f\left(\frac{2021}{2023}\right)+f\left(\frac{3}{2023}\right)+f\left(\frac{2020}{2023}\right).........f\left(\frac{1011}{2023}\right)+f\left(\frac{1012}{2023}\right)$

$=\underset{1}{\underbrace{f\left(\frac{1}{2023}\right)+f\left(\frac{2022}{2023}\right)}}+\underset{1}{\underbrace{f\left(\frac{2}{2023}\right)+f\left(\frac{2021}{2023}\right)}}+\underset{1}{\underbrace{f\left(\frac{3}{2023}\right)+f\left(\frac{2020}{2023}\right)}}.........+\underset{1}{\underbrace{f\left(\frac{1011}{2023}\right)+f\left(\frac{1012}{2023}\right)}}$

$=1+1+1.........1011 \text{times}=1011$