9. Given the sequence $\left\{a_{n}\right\}: a_{1}=7, \frac{a_{n+1}}{a_{n}}=a_{n}+2, n=1,2,3, \cdots$. Find the smallest positive integer $n$ such that $a_{n}>4^{2018}$.
$$
\begin{array}{l}
\frac{a_{n+1}}{a_{n}}=a_{n}+2 \Rightarrow a_{n+1}=a_{n}^{2}+2 a_{n} \Rightarrow a_{n+1}+1=\left(a_{n}+1\right)^{2} \\
\Rightarrow \ln \left(a_{n+1}+1\right)=2 \ln \left(a_{n}+1\right), \text { and } \ln \left(a_{1}+1\right)=\ln 8,
\end{array}
$$

Therefore, the sequence $\left\{\ln \left(a_{n}+1\right)\right\}$ is a geometric sequence with the first term $\ln 8$ and common ratio 2.
That is, $\ln \left(a_{n}+1\right)=2^{n-1} \ln 8 \Rightarrow a_{n}+1=8^{2^{n-1}} \Rightarrow a_{n}=8^{2^{n-1}}-1\left(n \in \mathbf{N}^{*}\right)$.
Thus, $a_{n}>4^{2018} \Rightarrow 2^{3 \cdot 2^{n-1}}>2^{4036}+1 \Rightarrow 3 \cdot 2^{n-1} \geqslant 4038$
$$
\Rightarrow 2^{n-1} \geqslant 1346 \Rightarrow n-1 \geqslant 11 \Rightarrow n \geqslant 12 \text {. }
$$

In summary, the smallest positive integer $n$ that satisfies $a_{n}>4^{2018}$ is $n=12$.