II. (20 points) Find the positive integer $n$ such that
$$
\left[\log _{3} 1\right]+\left[\log _{3} 2\right]+\cdots+\left[\log _{3} n\right]=2007 \text {, }
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$.
$$
\begin{array}{l}
{\left[\log _{3} 1\right]+\left[\log _{3} 2\right]=0,} \\
{\left[\log _{3} 3\right]+\left[\log _{3} 4\right]+\cdots+\left[\log _{3} 8\right]} \\
=6 \times 1=6, \\
{\left[\log _{3} 9\right]+\left[\log _{3} 10\right]+\cdots+\left[\log _{3} 26\right]} \\
=18 \times 2=36, \\
{\left[\log _{3} 27\right]+\left[\log _{3} 28\right]+\cdots+\left[\log _{3} 80\right]} \\
=54 \times 3=162, \\
{\left[\log _{3} 81\right]+\left[\log _{3} 82\right]+\cdots+\left[\log _{3} 242\right]} \\
=162 \times 4=648,
\end{array}
$$

and $\left[\log _{3} 243\right]+\left[\log _{3} 244\right]+\cdots+\left[\log _{3} 728\right]$
$$
=486 \times 5=2430>2007 \text {, }
$$

Therefore, $243<n<728$. Thus,
$$
6+36+162+648+(n-242) \times 5=2007 \text {. }
$$

Solving for $n$ gives $n=473$.
Hence, the positive integer $n$ that satisfies the condition is $n=473$.