Consider coins with positive real denominations not exceeding 1. Find the smallest $C>0$ such that the following holds: if we are given any 100 such coins with total value 50, then we can always split them into two stacks of 50 coins each such that the absolute difference between the total values of the two stacks is at most $C$.
The answer is $C=\frac{50}{51}$. The lower bound is obtained if we have 51 coins of value $\frac{1}{51}$ and 49 coins of value 1. (Alternatively, 51 coins of value $1-\frac{\varepsilon}{51}$ and 49 coins of value $\frac{\varepsilon}{49}$ works fine for $\varepsilon>0$.) We now present two (similar) proofs that this $C=\frac{50}{51}$ suffices. 【 First proof (original). Let $a_{1} \leq \cdots \leq a_{100}$ denote the values of the coins in ascending order. Since the 51 coins $a_{50}, \ldots, a_{100}$ are worth at least $51 a_{50}$, it follows that $a_{50} \leq \frac{50}{51}$; likewise $a_{51} \geq \frac{1}{51}$. We claim that choosing the stacks with coin values 
$$ a_{1}, a_{3}, \ldots, a_{49}, \quad a_{52}, a_{54}, \ldots, a_{100} $$
and 
$$ a_{2}, a_{4}, \ldots, a_{50}, \quad a_{51}, a_{53}, \ldots, a_{99} $$
works. Let $D$ denote the (possibly negative) difference between the two total values. Then 
$$ \begin{aligned} D & =\left(a_{1}-a_{2}\right)+\cdots+\left(a_{49}-a_{50}\right)-a_{51}+\left(a_{52}-a_{53}\right)+\cdots+\left(a_{98}-a_{99}\right)+a_{100} \\ & \leq 25 \cdot 0-\frac{1}{51}+24 \cdot 0+1=\frac{50}{51} \end{aligned} $$
Similarly, we have 
$$ \begin{aligned} D & =a_{1}+\left(a_{3}-a_{2}\right)+\cdots+\left(a_{49}-a_{48}\right)-a_{50}+\left(a_{52}-a_{51}\right)+\cdots+\left(a_{100}-a_{99}\right) \\ & \geq 0+24 \cdot 0-\frac{50}{51}+25 \cdot 0=-\frac{50}{51} \end{aligned} $$
It follows that $|D| \leq \frac{50}{51}$, as required.