1. We are given

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

Let

$$S=f\left(\frac1{2023}\right)+f\left(\frac2{2023}\right)+\cdots+f\left(\frac{2022}{2023}\right).$$

2. First, simplify the function a little:

$$2^{2x}=4^x.$$

So

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

3. Now compute $f(1-x)$:

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

Since $4^{1-x}=\frac{4}{4^x}$, we get

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

Multiply numerator and denominator by $4^x$:

f(1-x)=\frac{4}{4+2\cdot 4^x}=rac{2}{2+4^x}.

Also,

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

Hence,

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

4. In the required sum, terms pair up as:

$$f\left(\frac{k}{2023}\right)+f\left(1-\frac{k}{2023}\right) =f\left(\frac{k}{2023}\right)+f\left(\frac{2023-k}{2023}\right)=1.$$

5. The values of $k$ run from $1$ to $2022$. So the terms can be paired as:

$$\left(\frac1{2023},\frac{2022}{2023}\right), \left(\frac2{2023},\frac{2021}{2023}\right), \dots$$

There are

$$\frac{2022}{2}=1011$$

such pairs.

6. Each pair sums to $1$, so

$$S=1011.$$

7. Therefore the correct option is

$$\boxed{\text{D: }1011}.$$