I solved it.My main idea was something a little similar to 2003 KMO#2(?)

I believe this can help. Beatty sequence

PfTB 15.14

Actually quite easy. The answer is NO

Another simple proof: Such numbers do not exist. Suppose on the contrary that such number exists. Obviously $\alpha,\beta>1$ and that they are irrational. Let $x$ and $y$ be numbers such that $\frac{1}{\alpha}+\frac{1}{x}=\frac{1}{\beta}+\frac{1}{y}=1$, then by Beatty's theorem we have that the sequences $$B_x=\lfloor x\rfloor, \lfloor 2x\rfloor,...$$$$B_y=\lfloor y\rfloor, \lfloor 2y\rfloor,...$$satsify the condition $$B_x\cap B_y=\{2020,4040,...\},\quad B_x\cap B_y=\mathbb N$$CLAIM. $x<2$ Proof. Since $x$ is irrational, suppose on the contrary that $x>2$. Since $2020\in B_x$, let $mx=2020+d$ for some $0<d<1$. Let $k$ be a positive integer such that $\frac{1}{2}<kd<1$. Then $$2020\cdot(2k)+2>2mkx=2020\cdot(2k)+2kd>2020\cdot(2k)+1$$Since $x>2$ we have $$2020\cdot(2k)>(2mk-1)x$$Hence $2020\cdot (2k)$ does not belong to $B_x$, contradiction. $\blacksquare$ By symmetry $y<2$. However this is clearly impossible since $\lfloor x\rfloor=\lfloor y\rfloor=1$, so $1\in B_x\cap B_y$, contradiction.

Remark: This is just the result of Uspensky's Theorem, which states that $\mathbb{N}$ cannot be partitioned into $3$ or more Beatty sequences. In any case, here's a more reasonable proof that just uses Beatty's theorem, which states that given two irrationals $r, s$ satisfying $\tfrac{1}{r} + \tfrac 1s = 1$, then the beatty sequences generated by $r$ and $s$ partition $\mathbb{N}$. Let $r = \tfrac{\alpha}{\alpha - 1}$ and $s = \tfrac{\beta}{\beta - 1}$ so that $(\lfloor i\alpha \rfloor _{i = 1}^{\infty}, \lfloor ir \rfloor _{i = 1}^{\infty})$ and $(\lfloor i\beta \rfloor _{i = 1}^{\infty}, \lfloor is \rfloor _{i = 1}^{\infty})$ each partition $\mathbb{N}$ by Beatty. Assume that $\alpha, \beta$ satisfy the problem statement; then, we must have that $\lfloor ir \rfloor _{i = 1}^{\infty} \cap \lfloor is \rfloor _{i = 1}^{\infty}$ is precisely $\{2020i\}_{i = 1}^{\infty}$ and their union must include everything else. Clearly, $\alpha, \beta \geq 1$. We prove that in fact, they are $> 2$ too. Otherwise, that would imply either $r, s > 2$, WLOG $r$, and since $2020i \in \lfloor ir\rfloor _{i = 1}^{\infty}$, we must have $kr = 2020 + \epsilon$ for some $k$ and $\epsilon \in [0, 1)$. We find a contradiction, and in particular, show that some multiple of $2020$ is not in $\lfloor ir \rfloor _{i = 1}^{\infty}$. Indeed, let $n = \lfloor \tfrac{1}{\epsilon} \rfloor$, and note that\[(2kn)r = 2n(2020 + \epsilon) = 4040n + 2n\epsilon < 4040n + 2 \]and also\[4040n + 2n\epsilon = 4040n + 2\epsilon\lfloor \tfrac{1}{\epsilon}\rfloor > 4040n + 1\]but since $r > 2$, it follows that $(2kn - 1)r < 4040n$ hence $4040n$ does not actually belong in $\lfloor ir \rfloor _{i = 1}^{\infty}$, which is impossible, contradiction. Hence $\alpha, \beta > 2$. Then, $1$ does not appear, which is a contradiction, so the problem statement is impossible. $\blacksquare$