For clarity we write $p$ and $p+2$ for $20200$ and $20202$ whenever possible. The conditions are $a + c = p + 2$, $b + d = p$. By symmetry assume $b \le d$, then $1 \le b \le \frac{p}{2}$. In each sum $S = \frac{a}{b} + \frac{c}{d}$ replace $a$ and $c$ by their extremal values $a = 1, c = p + 1$ and $a = p + 1, c = 1$. In view of $a + c = p + 2$ and $b \le d$ comparison with $S$ shows respectively $(\frac{1}{b} + \frac{p+1}{d}) - S = (a-1)(\frac{1}{d} - \frac{1}{b}) \le 0$, $(\frac{p+1}{b} + \frac{1}{d}) - S = (c-1)(\frac{1}{b} - \frac{1}{d}) \ge 0$. Hence $\min S$ is attained with

a = 1, c = p+1, and $\max S$ with $a = p+1, c = 1$. Thus $\max S$ is the greatest value of $\frac{p+1}{b} + \frac{1}{p-b}$ where $1 \le b \le \frac{p}{2}$. It is straightforward that $b = 1$ yields a maximum. The result is $p+1 + \frac{1}{p-1}$ which is greater than $p+1$, while $b \ge 2$ implies $\frac{p+1}{b} + \frac{1}{p-b} \le \frac{p+1}{2} + 1 < p+1$. In particular $\max S = 20201 + \frac{1}{20199}$ for $p = 20200$, attained at $a = 20201, b = 1, c = 1, d = 20199$.

$$f(b) - f(b-1) = \frac{p(b^2 + b - p - 1)}{b(b-1)(p-b)(p-b+1)}$$
Because $b(b-1)(p-b)(p-b+1) > 0$ for $2 \le b \le \frac{p}{2}$, the sign of $f(b) - f(b-1)$ coincides with the sign of $b^2 + b - p - 1$. For $p = 20200$ this leads to the quadratic function $b^2 + b - 20201$. It has one negative root and one root between $141$ and $142$. Hence $b^2 + b - 20201 < 0$ for $2 \le b \le 141$ and $b^2 + b - 20201 > 0$ for $b \ge 142$. It follows that $f(1) > f(2) > \dots > f(141)$ and $f(141) < f(142) < \dots$, showing that $\min f$ is attained at $b = 141$ and equal to $\frac{1}{141} + \frac{20201}{20059}$. This is the minimum of $S$ under the given constraints, attained at $a = 1, b = 141, c = 20201, d = 20059$.