69th Belarusian Mathematical Olympiad

Answer: $\frac{A}{B}=4$.

(Solution by P. Verigo.) Let us solve the problem in more general case. Replace $50$ by an arbitrary positive integer $n>2$ of the form $n=4k+2$ and let the numbers initially written on the blackboard be $1,2,\dots ,n-1,n$.

For any $\ell$ numbers $a_1,a_2,\dots ,a_{\ell }$ written on the blackboard consider its characteristic defined by $$f(a_1,a_2,\dots ,a_{\ell })=\frac{1}{3}\left((a_1+a_2+\cdots +a_{\ell }{)}^3-a_1^3-a_2^3-\cdots -a_{\ell }^3\right).$$ Let $f_0$ be the characteristic of the initial numbers on the blackboard and $f_m$ be the characteristic of the numbers written after $m$ operations. It is easy to see that $$(a+b)(b+c)(c+a)=\frac{1}{3}\left((a+b+c{)}^3-a^3-b^3-c^3\right).$$ Hence at the $m^{\text{th}}$ operation Ann writes to her notebook the difference $f_{m-1}-f_m$. Therefore, the sum $X$ of all numbers written in the notebook after $\frac{n-2}{2}$ operations equals to $$f_0-f_{\frac{n-2}{2}}=\frac{1}{3}(S^3-(1^3+2^3+\cdots +n^3))-S\cdot x\cdot (S-x),$$ where $S=1+2+\cdots +n$, and the two numbers: $x$ and $S-x$ remain on the blackboard.

It is known that $1^3+2^3+\cdots +n^3=(1+2+\cdots +n{)}^2$, so $X=\frac{1}{3}(S^3-S^2)-S\cdot x(S-x)$.

Note that $S=(2k+1)(4k+3)$ is odd. Hence $\max x(S-x)=\frac{s-1}{2}\cdot \frac{s+1}{2}=\frac{s^2-1}{4}$ and $\min x(s-x)=1\cdot (s-1)=s-1$.

Therefore the maximum possible sum equals $$A=\frac{1}{3}(S^3-S^2)-S(S-1)=\frac{1}{3}S(S-1)(S-3),$$ the minimum possible sum equals $$B=\frac{1}{3}(S^3-S^2)-\frac{S(S^2-1)}{4}=\frac{1}{12}S(S-1)(S-3),$$ and clearly $\frac{A}{B}=4$.

It remains to verify that Ann can leave on the blackboard the numbers $x=\frac{s-1}{2}$ and $y=\frac{s+1}{2}$. It is sufficient to divide the set $\{1,2,\dots ,4k+2\}$ into two groups the sums of numbers in which differ by $1$. Consider the first group containing all odd numbers: $\{1,3,\dots ,4k+1\}$ and the second group containing all even numbers: $\{2,4,\dots ,4k+2\}$. The difference of their sums equals $2k+1$. Then we move the number $k$ from the second group to the first one if $k$ is even and the number $k+1$ from the second to the first if $k$ is odd. Thus the solution is finished.