Question 201, Find all positive integers $n$, such that $8^{n}+n$ is divisible by $2^{n}+n$.



Translating the above text into English, please retain the original text's line breaks and format, and output the translation result directly. 

However, it seems the instruction is redundant as the translation is already provided. Here is the translation again for clarity:

Question 201, Find all positive integers $n$, such that $8^{n}+n$ is divisible by $2^{n}+n$.
Question 201, Solution: Since $8^{n}+n=\left(2^{2 n}-n \cdot 2^{n}+n^{2}\right)\left(2^{n}+n\right)-\left(n-n^{3}\right)$, it follows that $2^{n}+n\left|8^{n}+n \Leftrightarrow 2^{n}+n\right| n^{3}-n$.
When $n^{3}-n=0 \Rightarrow n=1$, it is clear that $2^{n}+n \mid n^{3}-n$.
When $n^{3}-n>0$, then $2^{n}+n \leq n^{3}-n$.
Let $f(x)=\ln 2 \cdot x-3 \cdot \ln x$, then $f^{\prime}(x)=\ln 2-\frac{3}{x}>\frac{1}{2}-\frac{3}{x}$. Clearly, when $x \geq 6$, $f^{\prime}(x)>0$, so when $x \geq 6$, $f(x)=\ln 2 \cdot x-3 \cdot \ln x$ is monotonically increasing. And $f(10)=10 \cdot \ln 2-3 \cdot \ln 10=\ln \frac{1024}{1000}>0$, so when $x \geq 10$, $f(x)=\ln 2 \cdot x-3 \cdot \ln x>0 \Rightarrow 2^{x}>x^{3}$.
Thus, when $n \geq 10$, $2^{n}+n>2^{n}>n^{3}>n^{3}-n$. Therefore, $x \leq 9$.
Testing each value individually, the values of $n$ that satisfy the condition are $1, 2, 4, 6$.