$n(7a -3(n-1)!)= 2020$ Since $n\in\mathbb N$, we must have $n\mid 2020=2^{2}\cdot 5\cdot 101$. Thus, $n$ can only be $1,2,4,5,10,40,101,202,404,505,1010,2020$. $n=1$ means that $7a -3 = 2020\implies 7a=2023\implies a=289$. $n=2$ means that $14a -6 = 2020\implies 14a=2026\implies \text{no }a\in\mathbb N$. $n=4$ means that $28a -72 = 2020\implies 28=2092\implies \text{no }a\in\mathbb N$. $n=5$ means that $35a -360 = 2020\implies 35=2380\implies a=68$. Now, we have $n\geq 7$, therefore $7\mid n!$, hence $7\mid 7an+3n!$, but $2020\equiv 4\pmod{7}$, thus no $n\geq 7$. Answer. $\boxed{a=289 \text{ or } a=68}.$

If $n\geq7$ then LHS is divisible by $7$, but RHS is not a multiple of $7$ $n=6\Rightarrow 42a=2020+4180=6200$ No solution $n=5\Rightarrow 35a=2020+360=2380\Rightarrow a=68$ $n=4\Rightarrow 28a=2020+72=2092$ No solution $n=3\Rightarrow 21a=2020+18=2038$ No solution $n=2\Rightarrow 14a=2020+6=2026$ No solution $n=1\Rightarrow 7a=2023\Rightarrow a=289$ $n=0\Rightarrow 0=2023$