1. **Binary Representation of $\sqrt{2020}$ and $\sqrt{2021}$**:
Let the binary representation of $\sqrt{2020}$ be $\overline{\ldots a_0.a_1a_2\ldots}_2$ and the binary representation of $\sqrt{2021}$ be $\overline{\ldots b_0.b_1b_2\ldots}_2$.

2. **Expression for $V_n$**:
We have $V_n = \lfloor 2^n \sqrt{2020} \rfloor + \lfloor 2^n \sqrt{2021} \rfloor$.

3. **Condition for $V_n$ being even or odd**:
$V_n$ is even if and only if the sum of the fractional parts of $2^n \sqrt{2020}$ and $2^n \sqrt{2021}$ is less than 1. This happens if and only if $a_n = b_n$.

4. **Assumption of Finitely Many Odd $V_n$**:
Suppose there are finitely many odd $V_n$. This implies that for sufficiently large $n$, $a_n = b_n$.

5. **Implication of $a_n = b_n$ for Large $n$**:
If $a_n = b_n$ for all sufficiently large $n$, then the fractional parts of $2^n \sqrt{2020}$ and $2^n \sqrt{2021}$ are eventually the same. This implies that $\sqrt{2021} - \sqrt{2020}$ has a binary expansion that eventually terminates, making it a rational number.

6. Contradiction:
However, $\sqrt{2021} - \sqrt{2020}$ is not rational because if it were, then $(\sqrt{2021} - \sqrt{2020})^2 = 2021 + 2020 - 2\sqrt{2021 \cdot 2020}$ would imply that $2\sqrt{2021 \cdot 2020}$ is rational, which is not true since $2021 \cdot 2020$ is not a perfect square.

7. **Assumption of Finitely Many Even $V_n$**:
Suppose there are finitely many even $V_n$. This implies that for sufficiently large $n$, $a_n \neq b_n$, which means $a_n + b_n = 1$ since $a_n, b_n \in \{0, 1\}$.

8. **Implication of $a_n \neq b_n$ for Large $n$**:
If $a_n \neq b_n$ for all sufficiently large $n$, then the fractional parts of $2^n \sqrt{2020}$ and $2^n \sqrt{2021}$ are such that their sum is always 1. This implies that $\sqrt{2021} + \sqrt{2020}$ has a binary expansion that is eventually all 1's, which can be re-expressed as terminating (since $1_2 = 0.\overline{1}_2$), making it a rational number.

9. Contradiction:
However, $\sqrt{2021} + \sqrt{2020}$ is not rational because if it were, then $(\sqrt{2021} + \sqrt{2020})^2 = 2021 + 2020 + 2\sqrt{2021 \cdot 2020}$ would imply that $2\sqrt{2021 \cdot 2020}$ is rational, which is not true since $2021 \cdot 2020$ is not a perfect square.

10. Conclusion:
Since both assumptions lead to contradictions, it follows that there must be infinitely many odd $V_n$ and infinitely many even $V_n$.

$\blacksquare$