Compute the expected sum of elements in a subset of $\{1, 2, 3, . . . , 2020\}$ (including the empty set) chosen uniformly at random.
1. **Identify the problem**: We need to compute the expected sum of elements in a subset of $\{1, 2, 3, \ldots, 2020\}$ chosen uniformly at random, including the empty set.

2. **Use Linearity of Expectation**: The Linearity of Expectation states that the expected value of a sum of random variables is the sum of the expected values of those random variables. This property holds regardless of whether the random variables are independent.

3. **Probability of each element being in the subset**: For each element $i$ in the set $\{1, 2, 3, \ldots, 2020\}$, there is a $\frac{1}{2}$ probability that $i$ is included in a randomly chosen subset.

4. **Expected value for each element**: The expected contribution of each element $i$ to the sum of the subset is $i \cdot \frac{1}{2}$.

5. **Sum of expected values**: By the Linearity of Expectation, the expected sum of the elements in a subset is the sum of the expected contributions of each element:
   \[
   \mathbb{E}\left(\text{sum of elements in subset}\right) = \sum_{i=1}^{2020} i \cdot \frac{1}{2}
   \]

6. **Simplify the sum**: The sum of the first $n$ natural numbers is given by the formula:
   \[
   \sum_{i=1}^{n} i = \frac{n(n+1)}{2}
   \]
   For $n = 2020$, we have:
   \[
   \sum_{i=1}^{2020} i = \frac{2020 \cdot 2021}{2}
   \]

7. **Calculate the expected sum**: Using the above result, the expected sum of the elements in a subset is:
   \[
   \mathbb{E}\left(\text{sum of elements in subset}\right) = \frac{1}{2} \cdot \frac{2020 \cdot 2021}{2} = \frac{2020 \cdot 2021}{4}
   \]

8. **Final calculation**: Simplify the expression:
   \[
   \frac{2020 \cdot 2021}{4} = 505 \cdot 2021 = 1020605
   \]

The final answer is $\boxed{1020605}$.