Sure, here is the translated text:

```
II. (40 points) Given the set of positive integers
$$
S=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}
$$

satisfying that for any $S_{1} 、 S_{2} \subseteq S$, and $S_{1} \neq S_{2}$, we have
$$
\sum_{i \in S_{1}} i \neq \sum_{i \in S_{2}} j \text {. }
$$

Find the minimum value of $\sqrt{a_{1}}+\sqrt{a_{2}}+\cdots+\sqrt{a_{n}}$.
```
Let's assume $a_{1}<a_{2}<\cdots<a_{n}$, and denote
$$
T_{i}=\left\{a_{1}, a_{2}, \cdots, a_{i}\right\}(1 \leqslant i \leqslant n) .
$$

Then all the sums of the elements of the subsets of $T_{i}$ are distinct.
Hence, $a_{1}+a_{2}+\cdots+a_{i} \geqslant 2^{i}-1(1 \leqslant i \leqslant n)$.
Let $b_{1}=1, b_{2}=2, \cdots, b_{i}=2^{i-1}, \cdots, b_{n}=2^{n-1}$.
Then $b_{1}<b_{2}<\cdots<b_{n}$, and
$$
a_{1}+a_{2}+\cdots+a_{i} \geqslant b_{1}+b_{2}+\cdots+b_{i}(1 \leqslant i \leqslant n) \text {. }
$$

First, let's prove a lemma.
Lemma: Let $x, y \in \mathbf{R}_{+}$. Then
$$
\frac{x-y}{2 \sqrt{x}} \leqslant \sqrt{x}-\sqrt{y} \text {, }
$$

with equality if and only if $x=y$.
Proof: $\quad x+y \geqslant 2 \sqrt{x y} \Rightarrow x-y \leqslant 2 x-2 \sqrt{x y}$
$$
\Rightarrow x-y \leqslant 2 \sqrt{x}(\sqrt{x}-\sqrt{y}) \Rightarrow \frac{x-y}{2 \sqrt{x}} \leqslant \sqrt{x}-\sqrt{y} \text {. }
$$

Returning to the original problem.
Notice that,
$$
\begin{array}{l}
\sum_{i=1}^{n} \sqrt{a_{i}}-\sum_{i=1}^{n} \sqrt{b_{i}}=\sum_{i=1}^{n}\left(\sqrt{a_{i}}-\sqrt{b_{i}}\right) \\
\geqslant \sum_{i=1}^{n} \frac{a_{i}-b_{i}}{2 \sqrt{a_{i}}} \\
=\sum_{i=1}^{n}\left(\frac{1}{2 \sqrt{a_{i}}}-\frac{1}{2 \sqrt{a_{i+1}}}\right)\left(\sum_{j=1}^{i} a_{j}-\sum_{j=1}^{i} b_{j}\right) \geqslant 0 .
\end{array}
$$

Equality holds if and only if $a_{i}=b_{i}(1 \leqslant i \leqslant n)$.
$$
\text { And } S^{\prime}=\left\{b_{1}, b_{2}, b_{3} \cdots, b_{n}\right\}=\left\{1,2,4, \cdots, 2^{n-1}\right\}
$$

also satisfies the problem's requirements.
$$
\begin{array}{l}
\text { Hence }\left(\sum_{i=1}^{n} \sqrt{a_{i}}\right)_{\min }=\sum_{i=1}^{n} \sqrt{b_{i}} \\
=\left[(\sqrt{2})^{n}-1\right](\sqrt{2}+1) .
\end{array}
$$