Given a positive integer $n$. Find the largest real number $c$, such that: if the sum of the reciprocals of a set of integers greater than 1 (which can be the same) is less than $c$, then it is always possible to divide this set of numbers into no more than $n$ groups, so that the sum of the reciprocals of the numbers in each group is less than 1.
II. $c=\frac{n+1}{2}$.
Take this set of numbers as $a_{1}, a_{2}, \cdots, a_{4}$, if $a_{1}=a_{2}=\cdots=a_{s}=2$, then $c \leqslant \frac{n+1}{2}$.
We will prove the following by mathematical induction on $n$:
If $\sum_{i=1}^{s} \frac{1}{a_{i}}<\frac{n+1}{2}$, then $a_{1}, a_{2}, \cdots, a_{s}$ can be divided into no more than $n$ groups such that the sum of the reciprocals in each group is less than 1.
(1) When $n=1$, the conclusion holds.
(2) Assume the conclusion holds for $n-1$. Now consider the case for $n$.
Note that $a_{1} \geqslant 2$, i.e., $\frac{1}{a_{1}} \leqslant \frac{1}{2}$.
Let $t$ be the largest positive integer such that $\sum_{i=1}^{t-1} \frac{1}{a_{i}}<\frac{1}{2}$.
If $t=s+1$, then the conclusion holds.
If $t \leqslant s$, then $\sum_{i=1}^{t-1} \frac{1}{a_{i}}<\frac{1}{2} \leqslant \sum_{i=1}^{t} \frac{1}{a_{i}}$.
Also, $\sum_{i=1}^{t} \frac{1}{a_{i}}<\frac{1}{2}+\frac{1}{a_{i}} \leqslant 1$, so when $t=s$, the conclusion holds.
Now assume $t<s$, then $\sum_{i=t+1}^{i} \frac{1}{a_{i}}<\frac{n+1}{2}-\sum_{i=1}^{t} \frac{1}{a_{i}} \leqslant \frac{n}{2}$.
By the induction hypothesis, $a_{t+1}, \cdots, a$ can be divided into $n-1$ groups, each with a sum of reciprocals less than 1. Also, the sum of the reciprocals of $a_{1}, a_{2}, \cdots, a_{4}$ is less than 1, so $a_{1}, a_{2}, \cdots, a_{s}$ can be divided into $n$ groups, each with a sum of reciprocals less than 1.
By the principle of mathematical induction, the conclusion holds for all $n \geqslant 1$.