, 2020

Solution:
We first prove the inequality
$$((x+1)(y+1))^{2} \geq 4\left(x^{3}+y^{3}\right)$$
for real numbers $x, y \geq 0$ satisfying $|x-y| \leq 1$, with equality if and only if $\{x, y\}=\{0,1\}$ or $\{x, y\}=\{1,2\}$.
Indeed,
$$\begin{aligned} 4\left(x^{3}+y^{3}\right) & =4(x+y)\left(x^{2}-x y+y^{2}\right) \\ & \leq\left((x+y)+\left(x^{2}-x y+y^{2}\right)\right)^{2} \\ & =\left(x y+x+y+(x-y)^{2}\right)^{2} \\ & \leq(x y+x+y+1)^{2} \\ & =((x+1)(y+1))^{2}, \end{aligned}$$
where the first inequality follows by applying the AM-GM inequality on $x+y$ and $x^{2}-x y+y^{2}$ (which are clearly nonnegative). Equality holds in the first inequality precisely if $x+y=x^{2}-x y+y^{2}$ and in the second one if and only if $|x-y|=1$. Combining these equalities we have $x+y=(x-y)^{2}+x y=1+x y$ or $(x-1)(y-1)=0$, which yields the solutions $\{x, y\}=\{0,1\}$ or $\{x, y\}=\{1,2\}$.

Now, let $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ be any sequence satisfying conditions (i) and (ii) and let $\left(y_{1}, y_{2}, \ldots, y_{2020}\right)$ be any permutation of $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$. As $0 \leq \min \left(x_{i}, y_{i}\right) \leq \max \left(x_{i}, y_{i}\right) \leq \min \left(x_{i}, y_{i}\right)+1$, we can apply inequality (1) to the pair $\left(x_{i}, y_{i}\right)$ and sum over all $1 \leq i \leq 2020$ to conclude that
$$\sum_{i=1}^{2020}\left(\left(x_{i}+1\right)\left(y_{i}+1\right)\right)^{2} \geq 4 \sum_{i=1}^{2020}\left(x_{i}^{3}+y_{i}^{3}\right)=8 \sum_{i=1}^{2020} x_{i}^{3}$$
Therefore, in order to satisfy condition (iii), every inequality must be an equality. Hence, for every $1 \leq i \leq 2020$ we must have $\{x_{i}, y_{i}\}=\{0,1\}$ or $\{x_{i}, y_{i}\}=\{1,2\}$. By condition (ii), we see that either $\{x_{i}, y_{i}\}=\{0,1\}$ for all $i$ or $\{x_{i}, y_{i}\}=\{1,2\}$ for all $i$.

If $\{x_{i}, y_{i}\}=\{0,1\}$ for every $1 \leq i \leq 2020$, this implies that the sequences $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ and $\left(y_{1}, y_{2}, \ldots, y_{2020}\right)$ together have 2020 zeroes and 2020 ones. As $\left(y_{1}, y_{2}, \ldots, y_{2020}\right)$ is a permutation of $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ this implies that $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)=(0,0, \ldots, 0,1,1, \ldots, 1)$ with 1010 zeroes and 1010 ones. Conversely, note that this sequence satisfies conditions (i), (ii), and (iii) (in (iii), we take $\left(y_{1}, y_{2}, \ldots, y_{2020}\right)=\left(x_{2020}, x_{2019}, \ldots, x_{1}\right)$), showing that this sequence indeed works. The same reasoning holds for the case that $\{x_{i}, y_{i}\}=\{1,2\}$ for all $i$.

Therefore, the only solutions are:
$$(\underbrace{0,0, \ldots, 0}_{1010}, \underbrace{1,1, \ldots, 1}_{1010})$$
and
$$(\underbrace{1,1, \ldots, 1}_{1010}, \underbrace{2,2, \ldots, 2}_{1010})$$