2 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 may 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. (Chen Yong
Stroll to find $C_{\max }=\frac{n+1}{2}$.
On one hand, take a set of integers as $a_{1}=a_{2}=\cdots=a_{k}=2$, then it is easy to know that $C_{\max } \leqslant \frac{n+1}{2}$.

Below, we use mathematical induction on $n$ to prove that if a set of integers greater than 1, $a_{1}, \cdots, a_{k}$, satisfies
$$
\sum_{i=1}^{k} \frac{1}{a_{i}}<\frac{n+1}{2},
$$

then $a_{1}, a_{2}, \cdots, a_{k}$ can be divided into no more than $n$ groups, such that the sum of the reciprocals in each group is less than 1. This implies that $C_{\max } \geqslant \frac{n+1}{2}$, hence $C_{\max }=\frac{n+1}{2}$.
(1) When $n=1$, the conclusion holds.
(2) Assume the conclusion holds for $n-1$, now consider the case for $n$.
Notice 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 $t=k+1$, then the conclusion holds. If $t \leqslant k$, then $\sum_{i=1}^{t-1} \frac{1}{a_{i}}<\frac{1}{2} \leqslant \sum_{i=1}^{t} \frac{1}{a_{i}}$. Also, $\sum_{i=1}^{t} \frac{1}{a_{i}}<\frac{1}{2}+\frac{1}{a_{t+1}} \leqslant 1$. Hence, the conclusion holds when $t=k$.
Assume $t<k$, then $\sum_{i=t+1}^{k} \frac{1}{a_{i}}<\frac{n+1}{2}-\sum_{i=1}^{t} \frac{1}{a_{i}}<\frac{n}{2}$.
By the induction hypothesis, $a_{i+1}, \cdots, a_{k}$ can be divided into $n-1$ groups, each with a sum of reciprocals less than 1. Also, the sum of the reciprocals of $a_{1}, a_{2}, \cdots, a_{t}$ is less than 1. Therefore, $a_{1}, a_{2}, \cdots, a_{k}$ 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 $n \geqslant 1$. Proof completed.