$n$ is a positive integer. Let $a(n)$ be the smallest number for which $n\mid a(n)!$
Find all solutions of:$$\frac{a(n)}{n}=\frac{2}{3}$$
1. Suppose that $\frac{a(n)}{n} = \frac{2}{3}$. This implies that $a(n) = \frac{2}{3}n$.
2. Let $k$ be a positive integer such that $n = 3k$. Then $a(n) = 2k$.
3. By the definition of $a(n)$, $n \mid a(n)!$ and $n \nmid (a(n) - 1)!$. This translates to $3k \mid (2k)!$ and $3k \nmid (2k-1)!$.
4. We need to check the conditions for different values of $k$:
   - If $k \geq 4$, then $3k \leq 2k-1$. This implies that $3k \mid (2k-1)!$, which is a contradiction.
   - If $k = 2$, then $n = 6$ and $a(n) = 4$. We need to check if $6 \mid 4!$ and $6 \nmid 3!$:
     - $4! = 24$ and $6 \mid 24$, which is true.
     - $3! = 6$ and $6 \mid 6$, which is false. Hence, $k = 2$ is not a solution.
   - If $k = 1$, then $n = 3$ and $a(n) = 2$. We need to check if $3 \mid 2!$ and $3 \nmid 1!$:
     - $2! = 2$ and $3 \nmid 2$, which is true.
     - $1! = 1$ and $3 \nmid 1$, which is true. Hence, $k = 1$ is not a solution.
   - If $k = 3$, then $n = 9$ and $a(n) = 6$. We need to check if $9 \mid 6!$ and $9 \nmid 5!$:
     - $6! = 720$ and $9 \mid 720$, which is true.
     - $5! = 120$ and $9 \nmid 120$, which is true. Hence, $k = 3$ is a solution.

Therefore, the only solution is $k = 3$, which implies $n = 9$.

The final answer is $\boxed{n = 9}$.