$2 \cdot 50$ Let $n \geqslant 5$ be a natural number, and $n$ different natural numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the condition: for any two different non-empty subsets $A$ and $B$ of the set $S=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, the sum of all numbers in $A$ is not equal to the sum of all numbers in $B$. Under the above conditions, find the maximum value of the sum of reciprocals
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}
$$
[Solution] First, we prove the following lemma.

Lemma: Let $S=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ satisfy the conditions in the problem. Then for any natural number $k \leqslant n$, we have
$$
\sum_{i=1}^{k} a_{i} \geqslant 2^{k}-1 .
$$

Proof of the Lemma: If not, then
$$
\sum_{i=1}^{k} a_{i} \leqslant 2^{k}-2 \text {. }
$$

Since $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ has $2^{k}-1$ different non-empty subsets, their element sums all take values in $\left\{1,2, \cdots, 2^{k}-2\right\}$. By the pigeonhole principle, there must be two subsets with the same element sum, which contradicts the given condition. The lemma is proved.

Consider the expression
$$
\begin{array}{l}
1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n-1}}-\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right) \\
=\frac{a_{1}-1}{a_{1}}+\frac{a_{2}-2}{2 a_{2}}+\cdots+\frac{a_{n}-2^{n-1}}{2^{n-1} a_{n}} .
\end{array}
$$

Let $c_{i}=\frac{1}{2^{i-1} a_{i}}, d_{i}=a_{i}-2^{i-1}, D_{k}=\sum_{i=1}^{k} d_{i}$, then we have $c_{1}>c_{2}>\cdots>c_{n}$,
and
$$
\begin{aligned}
D_{k} & =\sum_{i=1}^{k} a_{i}-\left(1+2+2^{2}+\cdots+2^{k-1}\right) \\
& =\sum_{i=1}^{k} a_{i}-\left(2^{k}-1\right) \geqslant 0 .
\end{aligned}
$$

From (1) and (2), we have
$$
\begin{aligned}
& 1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n-1}}-\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right) \\
= & \sum_{i=1}^{n} c_{i} d_{i}=c_{1} D_{1}+C_{2}\left(D_{2}-D_{1}\right)+\cdots+c_{n}\left(D_{n}-D_{n-1}\right) \\
= & \left(c_{1}-c_{2}\right) D_{1}+\left(c_{2}-c_{3}\right) D_{2}+\cdots+\left(c_{n-1}-c_{n}\right) D_{n-1}+c_{n} D_{n} \geqslant 0 .
\end{aligned}
$$

Therefore,
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} \leqslant 1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n-1}}=2-\frac{1}{2^{n-1}} .
$$

On the other hand, it is easy to verify that the set $S_{0}=\left\{1,2,2^{2}, \cdots, 2^{n-1}\right\}$ satisfies the conditions in the problem and has
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}=1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n-1}}=2-\frac{1}{2^{n-1}} .
$$

In summary, the maximum value of the expression is $2-\frac{1}{2^{n-1}}$.