Answer. $k=2n-1$.

Solution 1. If $d=2n-1$ and $a_{1}=\cdots=a_{2n-1}=n/(2n-1)$, then each group in such a partition can contain at most one number, since $2n/(2n-1)>1$. Therefore $k \geqslant 2n-1$. It remains to show that a suitable partition into $2n-1$ groups always exists.
We proceed by induction on $d$. For $d \leqslant 2n-1$ the result is trivial. If $d \geqslant 2n$, then since
$$\left(a_{1}+a_{2}\right)+\ldots+\left(a_{2n-1}+a_{2n}\right) \leqslant n$$
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 $a_{1}, \ldots, a_{d}$.

Solution 2. We will show that it is even possible to split the sequence $a_{1}, \ldots, a_{d}$ into $2n-1$ contiguous groups so that the sum of the numbers in each group does not exceed $1$. Consider a segment $S$ of length $n$, and partition it into segments $S_{1}, \ldots, S_{d}$ of lengths $a_{1}, \ldots, a_{d}$, respectively, as shown below. Consider a second partition of $S$ into $n$ equal parts by $n-1$ "empty dots".

Assume that the $n-1$ empty dots are in segments $S_{i_{1}}, \ldots, S_{i_{n-1}}$. (If a dot is on the boundary of two segments, we choose the right segment). These $n-1$ segments are distinct because they have length at most $1$. Consider the partition:
$$\left\{a_{1}, \ldots, a_{i_{1}-1}\right\},\left\{a_{i_{1}}\right\},\left\{a_{i_{1}+1}, \ldots, a_{i_{2}-1}\right\},\left\{a_{i_{2}}\right\}, \ldots,\left\{a_{i_{n-1}}\right\},\left\{a_{i_{n-1}+1}, \ldots, a_{d}\right\}.$$
In the example above, this partition is $\left\{a_{1}, a_{2}\right\},\left\{a_{3}\right\},\left\{a_{4}, a_{5}\right\},\left\{a_{6}\right\}, \varnothing,\left\{a_{7}\right\},\left\{a_{8}, a_{9}, a_{10}\right\}$. We claim that in this partition, the sum of the numbers in each group is at most $1$.
For the sets $\left\{a_{i_{t}}\right\}$ this is obvious since $a_{i_{t}} \leqslant 1$. For the sets $\left\{a_{i_{t}+1}, \ldots, a_{i_{t+1}-1}\right\}$ this follows from the fact that the corresponding segments lie between two neighboring empty dots, or between an endpoint of $S$ and its nearest empty dot. Therefore the sum of their lengths cannot exceed $1$.

Solution 3. First put all numbers greater than $\frac{1}{2}$ in their own groups. Then, form the remaining groups as follows: For each group, add new $a_{i}$'s one at a time until their sum exceeds $\frac{1}{2}$. Since the last summand is at most $\frac{1}{2}$, this group has sum at most $1$. Continue this procedure until we have used all the $a_{i}$'s. Notice that the last group may have sum less than $\frac{1}{2}$. If the sum of the numbers in the last two groups is less than or equal to $1$, we merge them into one group. In the end we are left with $m$ groups. If $m=1$ we are done. Otherwise the first $m-2$ have sums greater than $\frac{1}{2}$ and the last two have total sum greater than $1$. Therefore $n>(m-2)/2+1$ so $m \leqslant 2n-1$ as desired.