The smallest number $n$ with this property is $4025$.

We first show $n \geq 4025$. For this we choose $d=4025$ as well as $a_{1}=\ldots=a_{4025}=\frac{2013}{4025}>\frac{1}{2}$. 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 needed here.

Now we show $n \leq 4025$. For this we perform a case distinction according to $d$.

For $d \leq 4025$, each $a_{i}$ receives its own subset. Thus all subsets, some of which may be empty, are disjoint and have element sums of at most $1$.

For $d>4025$, there must exist two numbers $a_{x}$ and $a_{y}$ with $a_{x}+a_{y} \leq 1$. Otherwise, already in the sum $\left(a_{1}+a_{2}\right)+\left(a_{3}+a_{4}\right)+\ldots+\left(a_{4025}+a_{4026}\right)$ every bracket would be greater than $1$ and the sum of all $a_{i}$ would be greater than $2013$, contradiction! Thus we can replace $a_{x}$ and $a_{y}$ by $a_{z}=a_{x}+a_{y}$ and obtain a set with $d-1$ elements satisfying all conditions. This step can be repeated until the replacement yields a set with $4025$ elements. For this set, and hence also for the original set, the desired partition exists. This proves everything.