Mathematical Olympiad Rioplatense

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 $f(1)>f(2)>\cdots >f(141)$ 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 $a=1,b=141,c=20201,d=20059$.