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
$$f(a_1, a_2, \dots, a_\ell) = \frac{1}{3}\left((a_1 + a_2 + \dots + a_\ell)^3 - a_1^3 - a_2^3 - \dots - a_\ell^3\right).$$
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
$$f_0 - f_{\frac{n-2}{2}} = \frac{1}{3}(S^3 - (1^3 + 2^3 + \dots + n^3)) - S \cdot x \cdot (S-x),$$
where $S = 1 + 2 + \dots + n$, and the two numbers: $x$ and $S-x$ remain on the blackboard.

It is known that $1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2$, 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.