7. Given a positive integer $n$. Find the smallest positive integer $k$, such that for any $d$ real numbers $a_{1}, a_{2}, \cdots, a_{d}$ satisfying
$$
a_{1}+a_{2}+\cdots+a_{d}=n\left(0 \leqslant a_{i} \leqslant 1, i=1,2, \cdots, n\right)
$$

they can be divided into no more than $k$ groups, and the sum of all numbers in each group does not exceed 1.
7. First prove: $k \geqslant 2 n-1$.

Take $d=2 n-1, a_{1}=a_{2}=\cdots=a_{2 n-1}=\frac{n}{2 n-1}$, it is easy to see that these numbers are all greater than $\frac{1}{2}$.
Therefore, any two of these numbers cannot be placed in the same group.
Thus, at least $2 n-1$ groups are needed.
Next, prove: $k \leqslant 2 n-1$.
Assume the partition with the minimum number of groups divides these $n$ numbers into $k$ groups $A_{1}, A_{2}, \cdots, A_{k}$, and let $\left\|A_{i}\right\|$ denote the sum of all elements in the set $A_{i}$.
Without loss of generality, assume $\left\|A_{1}\right\| \leqslant\left\|A_{2}\right\| \leqslant \cdots \leqslant\left\|A_{k}\right\|$.
If $k \geqslant 2 n$, by the principle of the average, we have
$$
\left\|A_{1}\right\|+\left\|A_{2}\right\| \leqslant \frac{2 n}{k} \leqslant 1 \text {. }
$$

Thus, at this point, $A_{1}$ and $A_{2}$ can be merged into one group, which contradicts the assumption that the number of groups is the minimum, hence $k \leqslant 2 n-1$.
Therefore, the smallest positive integer $k$ is $2 n-1$.