Let $n$ be a positive integer. Find the smallest integer $k$ with the following property: Given any real numbers $a_{1}, \ldots, a_{d}$ such that $a_{1}+a_{2}+\cdots+a_{d}=n$ and $0 \leqslant a_{i} \leqslant 1$ for $i=1,2, \ldots, d$, it is possible to partition these numbers into $k$ groups (some of which may be empty) such that the sum of the numbers in each group is at most 1.  (Poland)  Answer. $k=2 n-1$.

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If $d=2 n-1$ and $a_{1}=\cdots=a_{2 n-1}=n /(2 n-1)$, then each group in such a partition can contain at most one number, since $2 n /(2 n-1)>1$. Therefore $k \geqslant 2 n-1$. It remains to show that a suitable partition into $2 n-1$ groups always exists. We proceed by induction on $d$. For $d \leqslant 2 n-1$ the result is trivial. If $d \geqslant 2 n$, then since 
$$ \left(a_{1}+a_{2}\right)+\ldots+\left(a_{2 n-1}+a_{2 n}\right) \leqslant n $$
we may find two numbers $a_{i}, a_{i+1}$ such that $a_{i}+a_{i+1} \leqslant 1$. We "merge" these two numbers into one new number $a_{i}+a_{i+1}$. By the induction hypothesis, a suitable partition exists for the $d-1$ numbers $a_{1}, \ldots, a_{i-1}, a_{i}+a_{i+1}, a_{i+2}, \ldots, a_{d}$. This induces a suitable partition for $a_{1}, \ldots, a_{d}$.