2. Since $a(n)=\frac{2}{3} n$, then $3 \mid n$, let $n=3k$. If $k>3$, then $3k \mid k$! So $a(n) \leqslant k < \frac{2}{3} n$ is a contradiction. Also, since $a(3)=3 \neq \frac{2}{3} \times 3 \quad a(6)=3 \neq \frac{2}{3} \times 6 \quad a(9)=6=\frac{2}{3} \times 9$.
Therefore, $n=9$ is the solution.
2. If $n$ is a positive integer, $a(n)$ is the smallest positive integer such that $(a(n))!$ is divisible by $n$. Find all positive integers $n$ such that $\frac{a(n)}{n}=\frac{2}{3}$.
(2003 German Mathematical Olympiad)