The smallest number $n$ with this property is 4025.
We first show $n \geq 4025$. For this, we choose $d=4025$ and $a_{1}=\ldots=a_{4025}=\frac{2013}{4025}>\frac{1}{2}$. Then $a_{1}+\ldots+a_{4025}=2013$ and since $a_{i}+a_{j}=\frac{4026}{4025}>1$ for all $1 \leq i \neq j \leq 4025$, 4025 subsets are required.
Now we show $n \leq 4025$. For this, we perform a case distinction based on $d$.
For $d \leq 4025$, each $a_{i}$ gets its own subset. Thus, all subsets, some of which may be empty, are disjoint and have sums of at most 1.
For $d>4025$, there must exist two numbers $a_{x}$ and $a_{y}$ such that $a_{x}+a_{y} \leq 1$. Otherwise, in the sum $\left(a_{1}+a_{2}\right)+\left(a_{3}+a_{4}\right)+\ldots+\left(a_{4025}+a_{4026}\right)$, each parenthesis would be greater than 1, and the sum of all $a_{i}$ would be greater than 2013, a contradiction! Therefore, we can replace $a_{x}$ and $a_{y}$ with $a_{z}=a_{x}+a_{y}$ and obtain a set with $d-1$ elements that satisfies all conditions. This step can be repeated until the replacement yields a set with 4025 elements. For this set, and thus also for the original set, the desired partition exists. This completes the proof.

Note: Some of the $a_{i}$ may indeed be equal; two subsets are then disjoint if they do not contain the same element (with respect to the numbering). The problem statement allowed for many solution paths.
A natural number $n$ has the following property:
For arbitrary real numbers $a_{1}, a_{2}, \ldots, a_{d}$, which both satisfy $a_{1}+a_{2}+\ldots+a_{d}=2013$ and $0 \leq a_{i} \leq 1$ for $i=1,2, \ldots, d$, there exists a partition of the set of these real numbers into $n$ pairwise disjoint subsets (some of which may be empty), such that the sum of the numbers in each subset is at most 1.
Determine the smallest number $n$ with this property.