9. (16 points) Let $a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}\left(n \in \mathbf{Z}_{+}\right)$. Find the smallest positive real number $\lambda$, such that for any $n \geqslant 2$, we have
$$
a_{n}^{2}<\lambda \sum_{k=1}^{n} \frac{a_{k}}{k} .
$$
Sure, here is the translated text:

```
9. Notice that,
\[
\begin{array}{l}
a_{k}^{2}-a_{k-1}^{2}=\left(a_{k}-a_{k-1}\right)\left(a_{k}+a_{k-1}\right) \\
=\frac{1}{k}\left(2 a_{k}-\frac{1}{k}\right)(k \geqslant 2) .
\end{array}
\]

Then \(a_{n}^{2}-a_{1}^{2}=\sum_{k=2}^{n}\left(a_{k}^{2}-a_{k-1}^{2}\right)=2 \sum_{k=2}^{n} \frac{a_{k}}{k}-\sum_{k=2}^{n} \frac{1}{k^{2}}\)
\[
\begin{array}{l}
\Rightarrow a_{n}^{2}=2 \sum_{k=1}^{n} \frac{a_{k}}{k}-\sum_{k=1}^{n} \frac{1}{k^{2}} \\
\Rightarrow a_{n}^{2}\sum_{k=1}^{n} \frac{1}{k}.
\]
Taking \(n=2^{m}\), thus, when \(m\) is sufficiently large, we have
\[
\begin{array}{l}
\sum_{k=1}^{n} \frac{1}{k}=1+\sum_{i=0}^{m-1} \sum_{i=1}^{2^{i}} \frac{1}{2^{t}+s} \\
>1+\sum_{i=0}^{m-1} 2^{t} \times \frac{1}{2^{t+1}} \\
=1+\frac{m}{2}>\frac{2}{2-\lambda} .
\end{array}
\]

Therefore, equation (1) does not hold.
Thus, \(\lambda \geqslant 2\).
In conclusion, \(\lambda_{\min }=2\).
```