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.
Second proof (Evan Chen). Again we sort the coins in increasing order $0a_{i-1}+\frac{50}{51}$; obviously there is at most one such large gap.
Claim - If there is a large gap, it must be $a_{51}>a_{50}+\frac{50}{51}$.

Proof. If $i\frac{50}{51}$ and the sum $\sum_{1}^{100} a_{i}>50$ is too large. Conversely if $i>50$ then we get $a_{1}, \ldots, a_{i-1}<\frac{1}{51}$ and the sum $\sum_{1}^{100} a_{i}<1 / 51 \cdot 51+49$ is too small.

Now imagine starting with the coins $a_{1}, a_{3}, \ldots, a_{99}$, which have total value $S \leq 25$. We replace $a_{1}$ by $a_{2}$, then $a_{3}$ by $a_{4}$, and so on, until we replace $a_{99}$ by $a_{100}$. At the end of the process we have $S \geq 25$. Moreover, since we did not cross a large gap at any point, the quantity $S$ changed by at most $C=\frac{50}{51}$ at each step. So at some point in the process we need to have $25-C / 2 \leq S \leq 25+C / 2$, which proves $C$ works.