solution?

Solved with JustinLee2017 The knight can guarantee himself a freedom in at most $\boxed{25}$ days. Construction: Start by having the knight split the coins into a pile of 75 and a pile of 25. Each day, the knight takes one coin from the larger pile and puts it in the smaller pile. Proof of construction: Suppose there are $x$ magical coins in the 75 pile on the first day. Thus, there are $75-x$ normal coins in the 75 pile, $50-x$ magical coins in the 25 pile, and $x-25$ normal coins in the 25 pile. Note that $75-x \leq 50$ and so $x \geq 25$. The knight takes at least $$\text{min}\left(\frac{|(x)-(50-x)|}{2}, \frac{|(75-x)-(x-25)|}{2}\right)$$days to finish, and thus takes at most $$\left(\frac{|(x)-(50-x)|}{2} - 1\right) + \left(\frac{(75-x)-(x-25)|}{2} - 1\right) +1$$days to finish. Since $x \geq 25$, we have that $x \leq 50-x$ and $x-25 \leq 75-x$, and we can rewrite this as $$\left(\frac{50-2x}{2}-1\right) + \left(\frac{100-2x}{2}-1\right) + 1 = 74-2x \leq 24$$ Therefore, using this method, the knight can guarantee his freedom in $25$ days.