Interestingly, the answer is $\boxed{4}$. The key idea is the following identity: $$3(a+b)(b+c)(c+a) = (a+b+c)^3 - a^3 - b^3 - c^3. \qquad (1)$$ If we then let $x, y$ be the last two remaining numbers on the board, then Ann's sum is simply: $$\frac{x^3 + y^3 - 1^3 - 2^3 - 3^3 - \cdots - 50^3}{3}.$$ Since $x, y > 0$ and $x+y = \frac12 \cdot 50 \cdot 51 = 1275,$ it's clear that $637^3 + 638^3 \le x^3 + y^3 \le 1274^3 + 1.$ Furthermore, it's easy to see that it is indeed possible for Ann to force $(x, y) = (1, 1274)$ or $(x, y) = (637, 638).$ Hence, our answer is: $$\frac{A}{B} = \frac{1+ 1274^3 - 1^3 -2^3 - 3^3 - \cdots - 50^3}{637^3 + 638^3 - 1^3 - 2^3 - \cdots - 50^3} = \boxed{4}.$$ $\square$