Fall AMC 10 B, 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 $p = 2n + 1 = 4041$.