GPT答：

7i24.cn

Let the three consecutive positive integers be $n-1$, $n$, and $n+1$. Then we have $(n-1)n(n+1) = 2020n$. Simplifying, we get $(n-1)n(n+1) = 2^2 \cdot 5 \cdot 101 \cdot n$. Since $\gcd(n-1,n)=\gcd(n,n+1)=\gcd(n+1,n-1)=\gcd(2,n-1)$, one of the factors on the left-hand side must be divisible by $4$ and another factor must be divisible by $5$.

If $(n-1)$ is divisible by $4$, then either $(n-3)$ or $(n+3)$ is also divisible by $4$. However, since $\gcd(2,n\pm3)=\gcd(6,n\pm3)=\cdots=2$, none of these numbers can be equal to a multiple of any other factor. Thus, it follows that either $(n)(n+1)$ or $(n)( n -  11 )$ is a multiple of four.

If neither of them are multiples of four, then one must be a multiple of five. This means that either: 
\[ ( n -   5 ) ( n +   6 ) =    n^2+n -30     \] 
or
\[ ( n -   6 ) ( n +   5 ) =    n^2-n -30     \]
is a multiple of five. Therefore,
\[ |{ {    {      {       } } }_{ {(a)} } }\;{{ {{    {      {       } }} }}_{ {(b)} }}|=\left|\frac{n^{2} + n}{20}\right|,\;\;\; |{ {\;{{ {{    {      {       } }} }}}}_{ {(c)}} }\;{{ {{    {      {
} }}}}_{ {(d)}} }\}|=\left|\frac{n^{2}-9}{20}\right|. \]
Since at least one out two pairs ${(a,b),(c,d)}$ has an integer solution for ${m}$ such that ${m(m +   d)}/{20}=k,$ where ${k}$ is an integer,
we find that the smallest possible value for ${(a,b,c,d,k,m)}$
is ${(9,-10,-8,7,-18,-10)}.$ Thus our answer is $\boxed{010}.$