1. The answer is $c=\frac{1000}{91}=11-\frac{11}{1001}$. Clearly, if $c^{\prime}$ works, so does any $c>c^{\prime}$. First we prove that $c=11-\frac{11}{1001}$ is good.

We start with 100 empty boxes. First, we consider only the coins that individually value more than $\frac{1000}{1001}$. As their sum cannot overpass 1000, we deduce that there are at most 1000 such coins. Thus we are able to put (at most) 10 such coins in each of the 100 boxes. Everything so far is all right: $10 \cdot \frac{1000}{1001} < 11 - \frac{11}{1001}$.

for all $i=1,2, \ldots, 100$. Then

$$
x_{1}+x_{2}+\cdots+x_{100}+100 x > 100 \cdot\left(11-\frac{11}{1001}\right) .
$$

But since $1000 \geqslant x_{1}+x_{2}+\cdots+x_{100}+x$ and $\frac{1000}{1001} \geqslant x$ we obtain the contradiction

$$
1000+99 \cdot \frac{1000}{1001} > 100 \cdot\left(11-\frac{11}{1001}\right) \Longleftrightarrow 1000 \cdot \frac{1100}{1001} > 100 \cdot 11 \cdot \frac{1000}{1001} .
$$

Thus the algorithm does not fail and since we have finitely many coins, we will eventually reach to a happy end.

Now we show that $c=11-11 \alpha$, with $1>\alpha>\frac{1}{1001}$ does not work.
Take $r \in\left[\frac{1}{1001}, \alpha\right)$ and let $n=\left\lfloor\frac{1000}{1-r}\right\rfloor$. Since $r \geqslant \frac{1}{1001}$, then $\frac{1000}{1-r} \geqslant 1001$, therefore $n \geqslant 1001$.
Now take $n$ coins each of value $1-r$. Their sum is $n(1-r) \leqslant \frac{1000}{1-r} \cdot(1-r)=1000$. Now, no matter how we place them in 100 boxes, as $n \geqslant 1001$, there exist 11 coins in the same box. But $11(1-r)=11-11 r>11-11 \alpha$, so the constant $c=11-11 \alpha$ indeed does not work.
In an exotic country, the National Bank issues coins that can take any value in the interval $[0,1]$. Find the smallest constant $c>0$ such that the following holds, no matter the situation in that country:

Any citizen of the exotic country that has a finite number of coins, with a total value of no more than 1000, can split those coins into 100 boxes, such that the total value inside each box is at most $c$.

## Proposed by Romania