Consider the integer $N = 999988887777\ldots 0000$ where each digits $0,1,2,\ldots,9$ occur 4 times. Clearly, $N$ is divisible by both $1000$ and $1111$. Since $1000$ is divisible by $20$ and $1111$ is divisible by $101$, then $N$ must be divisible by $2020$.

Well How did you think of this number

ok I got it

consider the following number $N=123456789123456789.....0..0$($k$ times 123456789 and $k$ times $0$) So,$N=10^k*123456789(1+10^9+10^{2*9}+\cdots+10^{(k-1)*9})$.We can easily make the term inside the first bracket divisible by $202$ taking $k=\phi(202)$.Hence we are done.

themathematicalmantra wrote: ok I got it Please share how

Moreover, what is the minimum of such numbers? (minimum positive integer that is divisible by $2020$ and has equal numbers of digits $0, 1, 2, . . . , 9$ ?)

Observe that $\overline{abab}=101\cdot \overline{ab}$. Hence, for example integer $N=98987676545431312020$ is divisible by $101$. Also, notice that $4\mid N$, since the last two numbers are divisible by $4$ and $5\mid N$, because it ends with the digit $0$.

10132325454767698980 works, I think