Let $P$ be a polynomial with integer coefficients, and let $n$ be the cardinality of the set

$$
\{k \in \mathbb{Z}: k P(k)=2020\}
$$

What are the possible values of $n$?
Consider the polynomial $Q(X)=X P(X)-2020$, which has $n$ distinct integer roots $a_{1}, a_{2}, \ldots, a_{n}$. Then $Q(X)$ is divisible by the polynomial $R(X)=\left(X-a_{1}\right)\left(X-a_{2}\right) \cdots\left(X-a_{n}\right)$. This means in particular that $a_{1}, a_{2}, \ldots, a_{n}$ are distinct divisors of $R(0)$, which itself divides $Q(0)=-2020=-2^{2} \times 5 \times 101$. We deduce that $n \leqslant 6$.

Conversely, if one wishes to have $n=0$, it suffices to choose $P(X)=0$. Then, if one wishes to achieve any $n$ between 1 and 6, one starts by constructing the integers $a_{1}, a_{2}, \ldots, a_{n-1}$ as the first $n-1$ terms of the sequence $-1,1,-2,2,5$, then one sets

$$
a_{n}=(-1)^{n+1} \frac{2020}{a_{1} a_{2} \cdots a_{n-1}}
$$

It is easily verified that $a_{n}$ is also an integer, and then one can directly choose

$$
P(X)=\frac{\left(X-a_{1}\right)\left(X-a_{2}\right) \cdots\left(X-a_{n}\right)+2020}{X}
$$