, 2021

The given expression equals
$$\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021^2 - 2020^2}{2020 \cdot 2021}$$

$$\begin{aligned} &= \frac{(2021 + 2020)(2021 - 2020)}{2020 \cdot 2021} \\ &= \frac{4041}{2020 \cdot 2021}. \end{aligned}$$
Because $4041 - 2 \cdot 2020 = 1$, it follows that $4041$ and $2020$ cannot have a common divisor greater than $1$. Similarly, because $4041 - 2 \cdot 2021 = -1$, it follows that $4041$ and $2021$ cannot have a common divisor greater than $1$. Hence $\frac{4041}{2020 \cdot 2021}$ is in simplest terms, and the requested numerator is $4041$. Let $n = 2020$. Then the given fraction equals
$$\begin{aligned} \frac{n+1}{n} - \frac{n}{n+1} &= \frac{(n+1)^2}{n(n+1)} - \frac{n^2}{n(n+1)} \\ &= \frac{n^2 + 2n + 1}{n(n+1)} - \frac{n^2}{n(n+1)} \\ &= \frac{2n + 1}{n(n+1)}. \end{aligned}$$
$$$$
Note that if $d$ is a divisor of both $a$ and $a+b$, then $d$ is also a divisor of their difference, $b$. Because $n$ and $n+1$ have no common divisors greater than $1$, it follows that $2n+1$ can have no common divisors greater than $1$ with either $n$ or $n+1$. Thus $p = 2n + 1 = 4041$.