Let's assume $1 \leqslant a_{1}<a_{2}<\cdots<a_{n}$. First, we prove that for any $i(1 \leqslant i \leqslant n)$, it must hold that
$$\sum_{j=1}^{i} a_{j} \geqslant 2^{i}-1 \text {. }$$

Since the $2^{i}-1$ non-empty subsets of $\left\{a_{1}, a_{2}, \cdots, a_{i}\right\}$ have distinct sums, and these sums are all positive integers not exceeding $\sum_{j=1}^{i} a_{j}$, we have
$$\sum_{j=1}^{i} a_{j} \geqslant 2^{i}-1 .$$

Let $A_{i}=\sum_{j=1}^{i} a_{j}-\left(2^{i}-1\right)$. Then
$$A_{i} \geqslant 0 \text { and } A_{i}-A_{i-1}=a_{i}-2^{i-1}(1 \leqslant i \leqslant n) \text {. }$$

Let $A_{0}=0$. Then
$$\begin{array}{l} \sum_{i=1}^{n} \sqrt{a_{i}}-\sum_{i=1}^{n} \sqrt{2^{i-1}}=\sum_{i=1}^{n}\left(\sqrt{a_{i}}-\sqrt{2^{i-1}}\right) \\ =\sum_{i=1}^{n} \frac{a_{i}-2^{i-1}}{\sqrt{a_{i}}+\sqrt{2^{i-1}}}=\sum_{i=1}^{n} \frac{A_{i}-A_{i-1}}{\sqrt{a_{i}}+\sqrt{2^{i-1}}} \\ =\sum_{i=1}^{n-1} A_{i}\left(\frac{1}{\sqrt{a_{i}}+\sqrt{2^{i-1}}}-\frac{1}{\sqrt{a_{i+1}}+\sqrt{2^{i}}}\right)+ \\ \frac{A_{n}}{\sqrt{a_{n}}+\sqrt{2^{n-1}}} \\ \geqslant 0 \\ \Rightarrow \sum_{i=1}^{n} \sqrt{a_{i}} \geqslant \sum_{i=1}^{n} \sqrt{2^{i-1}} \\ \quad=(\sqrt{2}+1)\left(\sqrt{2^{n}}-1\right) . \end{array}$$
$$\text { When } A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}=\left\{1,2, \cdots, 2^{n-1}\right\}$$

the equality holds.
Thus, the minimum value of $\sum_{i=1}^{n} \sqrt{a_{i}}$ is $(\sqrt{2}+1)\left(\sqrt{2^{n}}-1\right)$.