Let $n$ be a strictly positive integer. Find the smallest integer $k$ having the following property: for all real numbers $a_{1}, \ldots, a_{d}$ satisfying $a_{1}+a_{2}+\cdots+a_{d}=n$ and $0 \leqslant a_{i} \leqslant 1$ for $i=1,2, \ldots, d$, it is possible to group the numbers into $k$ packets (possibly empty) such that the sum of the numbers in each packet is $\leqslant 1$.
$\triangleright$ The reals $a_{1}=\ldots=a_{2 n-1}=\frac{n}{2 n-1}$ belong to the interval $\left.] \frac{1}{2}, 1\right]$ and their sum is $n$. Since each real is strictly greater than $\frac{1}{2}$, each package contains at most one of these $2 n-1$ reals (because the sum of two of them is strictly greater than 1): therefore, at least $2 n-1$ packages are needed.
$\triangleright$ Suppose we have partitioned our set of reals $a_{1}, \ldots, a_{d}$ into $k \geq 2 n$ packages and let $\alpha_{1} \leq \ldots \leq \alpha_{k}$ be the sums associated with these packages, arranged in ascending order. Then, by grouping these numbers two by two, we get

$$
\begin{aligned}
n & =\alpha_{1}+\alpha_{2}+\ldots+\alpha_{k} \\
& =\left(\alpha_{1}+\alpha_{2}\right)+\left(\alpha_{3}+\alpha_{4}\right)+\ldots \\
& \geq\left\lfloor\frac{k}{2}\right\rfloor\left(\alpha_{1}+\alpha_{2}\right) \geq n\left(\alpha_{1}+\alpha_{2}\right)
\end{aligned}
$$

Thus, $\alpha_{1}+\alpha_{2} \leq 1$ and we can group the two corresponding packages into a new package with a total sum less than or equal to 1, and we are reduced to $k-1$ packages.
$\triangleright$ In conclusion, it is necessary to have $2 n-1$ packages, and if we have partitioned into more than $2 n$ packages, we can proceed with a grouping and reduce the number of necessary packages. The value sought is therefore $2 n-1$.