To determine the smallest integer $k$ that allows partitioning the numbers $a_1, a_2, \ldots, a_d$ into groups such that the sum of numbers in each group does not exceed 1, we start by analyzing the given constraints:

1. $a_1 + a_2 + \cdots + a_d = n$.
2. $0 \le a_i \le 1$ for $i = 1, 2, \ldots, d$.

### Objective
We want to partition these numbers into $k$ groups such that the sum in each group is at most 1.

### Analysis
Consider the worst-case scenario where each $a_i$ is as small as possible but still greater than zero. This will maximize the number of groups needed to cover all numbers.

1. Each $a_i$ is close to 1, the maximum permissible value, which reduces the sum more effectively per group.
2. In the extreme case, achieving as close to zero for most values, consider an example: $a_1 = a_2 = \ldots = a_{2n-1} = \frac{n}{2n-1}$. This creates a large number of values that are less than 1 but together sum to n.

### Calculating $k$
If each group’s sum is strictly less than or equal to 1:

- The minimal effective partition size ensures that each possible sum $\frac{n}{2n-1}$ or close not exceeded 1 in any group.
- We see that combining maximum pairs gives exactly integer partisans.

Therefore, to satisfy this, for each set value in structure $\lceil (2n-1)/2 \rceil = n$ grouping pattern, there must be at least:

$$k = 2n - 1$$

as it aligns meeting partition reliably with aggregate $\frac{n}{2n-1}$ in every group and not exceeding maximal individual sum constraint.

### Conclusion
Thus, the smallest integer $k$ that fulfills the condition is:
$$\boxed{2n-1}$$
This satisfies both our summation and group partition requirements.