Example 8 (2003 Bulgaria National Mathematical Olympiad) Find all real numbers $a$ that satisfy the condition $4[a n]=n+[a[a n]]$, where $n$ is any positive integer.

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

(Note: The provided text is already in English, so no translation is needed for this specific case.)
If $a \leqslant 0$, then $[a n] \leqslant 0, a[a n] \geqslant 0$.
Thus, $n+[a[a n]] \geqslant n+0>0 \geqslant 4[a n]$, contradiction!
Therefore, $a>0$.
Then, $4(a n-1)n+a(a n-1)-1$,
so $4 a>1+a^{2}-\frac{a+1}{n}$.
And $4 a<1+a^{2}+\frac{4}{n}$, i.e., $1+a^{2}-\frac{a+1}{n}<4 a<1+a^{2}+\frac{4}{n}$.
Let $n \rightarrow+\infty$, we get $1+a^{2}=4 a$,
so $a=2-\sqrt{3}$ or $a=2+\sqrt{3}$.
Let $n=1$, then $4[a]=1+[a[a]]$, and $a=2-\sqrt{3}$ does not satisfy $4[a]=1+[a[a]]$. Hence, $a \neq 2-\sqrt{3}$.
When $a=2+\sqrt{3}$, for any $n \in \mathbf{N}_{+}$, let $\left[\frac{n}{a}\right]=b,\left\{\frac{n}{a}\right\}=c$.
Then, from $1+a^{2}=4 a$, we know $a=4-\frac{1}{a}$, so
$$
\begin{array}{l}
n+[a[a n]]=\left[n+a\left[4 n-\frac{n}{a}\right]\right] \\
=[n+a(4 n-b-1)] \\
=[a(4 n-1+c)]=\left[\left(4-\frac{1}{a}\right)(4 n-1+c)\right] \\
=\left[4(4 n-1)-4\left(\frac{n}{a}-c\right)+\frac{1-c}{a}\right] \\
=4(4 n-1-b)=4\left[4 n-\frac{n}{a}\right]=4[a n] .
\end{array}
$$

This means $a=2+\sqrt{3}$ satisfies the condition.
Therefore, $a=2+\sqrt{3}$ is the only solution.