Example 8 For a positive integer $n$, find the smallest integer $k$ such that for any given real numbers $a_{1}, a_{2}, \cdots, a_{d}$, we have
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{d}=n, \\
0 \leqslant a_{i} \leqslant 1(i=1,2, \cdots, d),
\end{array}
$$

these real numbers can be divided into $k$ groups (allowing some groups to be empty sets) such that the sum in each group is at most $1 .{ }^{[3]}$
【Analysis】When $a_{1}=a_{2}=\cdots=a_{2 n-1}=\frac{n}{2 n-1}$, at least $2 n-1$ groups are needed.
Next, we prove that $2 n-1$ groups are definitely feasible.
To use as few sets as possible, we might as well number the sets and greedily place each real number into the set with the smallest possible number, assuming that $m$ sets are used at this point, and the sums of the real numbers in each set are $g_{1}, g_{2}, \cdots, g_{m}$. Then
$g_{i}+g_{j}>1(1 \leqslant i < j \leqslant m)$.
Therefore, $m \leqslant 2 n-1$.