Question 215, Let the sum of all positive divisors of the positive integer $\mathrm{n}$ be $\sigma(\mathrm{n})$, find the value: $\left[\sqrt[3]{\sum_{\mathrm{n}=1}^{2020} \frac{\sigma(\mathrm{n}}{\mathrm{n}}}\right]$, here $[\mathrm{x}]$ represents the greatest integer not exceeding $x$.

保留源文本的换行和格式，翻译结果如下：

Question 215, Let the sum of all positive divisors of the positive integer $\mathrm{n}$ be $\sigma(\mathrm{n})$, find the value: $\left[\sqrt[3]{\sum_{\mathrm{n}=1}^{2020} \frac{\sigma(\mathrm{n}}{\mathrm{n}}}\right]$, here $[\mathrm{x}]$ represents the greatest integer not exceeding $x$.
Question 215, Solution: Note that
$$
\begin{array}{l}
\sum_{\mathrm{n}=1}^{2020} \frac{\sigma(\mathrm{n})}{\mathrm{n}}=\sum_{\mathrm{n}=1}^{2020} \frac{\sum_{\mathrm{d} \mid \mathrm{n}} \mathrm{d}}{\mathrm{n}}=\sum_{\mathrm{n}=1}^{2020} \sum_{\mathrm{d} \mid \mathrm{n}} \frac{1}{\mathrm{~d}} \\
=\sum_{\mathrm{d}=1}^{2020} \sum_{1 \leq \mathrm{n} \leq 2020, \mathrm{d} \mid \mathrm{n}} \frac{1}{\mathrm{~d}}=\sum_{\mathrm{d}=1}^{2020} \frac{1}{\mathrm{~d}} \cdot\left[\frac{2020}{\mathrm{~d}}\right] \\
\end{array}
$$

On the one hand, according to (*) we know:
$$
\begin{array}{l}
\sum_{n=1}^{2020} \frac{\sigma(n)}{n} \cdot 2020 + \frac{1}{2} \times 1010 + \frac{1}{3} \times 673 + \frac{1}{4} \times 505 > 14^{3} \\
\text { Therefore, }\left[\sqrt[3]{\sum_{n=1}^{2020} \frac{\sigma(n)}{n}}\right]=14 .
\end{array}
$$