Let's see how, as a result of Ann's actions, the sum of the cubes of the numbers written on the board changes. The identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ implies that after each action of Ann, the sum of the cubes of all numbers written on the board increases by the number which is three times greater than the one that Ann writes down to her notebook. Therefore the sum of all numbers written in the notebook is equal to the difference between the cube of the last number remaining on the board and the sum of the cubes of all numbers written on the board initially. The number that remains on the board is equal to the sum of all numbers written on it initially which means that
$$3S = (1 + 2 + \dots + 50)^3 - (1^3 + 2^3 + \dots + 50^3).$$
We use the well-known identity
$$1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2 = \left(\frac{n(n+1)}{2}\right)^2,$$
which is easy to prove by induction. Since $\frac{50 \cdot 51}{2} = 1275$, then
$$3S = 1275^3 - 1275^2 = 1275^2 \cdot 1274 = 3 \cdot 425 \cdot 1274 \cdot 1275.$$
So the number $S$ is equal to $425 \cdot 1274 \cdot 1275 = 690348750$.