Let $n\in \mathbb{Z}_{> 0}$. The  set $S$ contains all positive integers written in decimal form that simultaneously satisfy the following conditions:
[list=1][*] each element of $S$ has exactly $n$ digits;
[*] each element of $S$ is divisible by $3$;
[*] each element of $S$ has all its digits from the set $\{3,5,7,9\}$ [/list]
Find $\mid S\mid$
1. **Define the problem and initial conditions:**
   - We need to find the number of $n$-digit numbers composed of the digits $\{3, 5, 7, 9\}$ that are divisible by $3$.
   - Let $a_n$ be the number of such $n$-digit numbers.

2. **Establish the recursion:**
   - Consider the number of $n$-digit numbers using the digits $\{3, 5, 7, 9\}$. There are $4^n$ such numbers.
   - We need to count how many of these numbers are divisible by $3$.
   - A number is divisible by $3$ if the sum of its digits is divisible by $3$.
   - Let $b_n$ be the number of $n$-digit numbers that are not divisible by $3$. Then, $a_n = 4^n - b_n$.

3. **Analyze the digits:**
   - The digits $\{3, 5, 7, 9\}$ have the following properties with respect to modulo $3$:
     - $3 \equiv 0 \pmod{3}$
     - $5 \equiv 2 \pmod{3}$
     - $7 \equiv 1 \pmod{3}$
     - $9 \equiv 0 \pmod{3}$
   - Therefore, the sum of the digits of any number formed by these digits can be $0, 1, 2 \pmod{3}$.

4. **Formulate the recursion:**
   - For $n=1$, the possible digits are $\{3, 5, 7, 9\}$.
     - Only $3$ and $9$ are divisible by $3$, so $a_1 = 2$.
   - For $n+1$ digits, consider the last digit:
     - If the last digit is $3$ or $9$, the number of $n$-digit numbers that are divisible by $3$ is $a_n$.
     - If the last digit is $5$ or $7$, the number of $n$-digit numbers that are not divisible by $3$ is $b_n$.
   - Therefore, the recursion can be written as:
     \[
     a_{n+1} = 2a_n + 2b_n
     \]
   - Since $b_n = 4^n - a_n$, we can substitute:
     \[
     a_{n+1} = 2a_n + 2(4^n - a_n) = 2a_n + 2 \cdot 4^n - 2a_n = 2 \cdot 4^n
     \]
     \[
     a_{n+1} = 4^n + a_n
     \]

5. **Solve the recursion:**
   - We have the initial condition $a_1 = 2$.
   - The recursion is $a_{n+1} = 4^n + a_n$.
   - We can solve this by summing the series:
     \[
     a_n = 2 + 4^1 + 4^2 + \cdots + 4^{n-1}
     \]
   - This is a geometric series with the first term $2$ and common ratio $4$:
     \[
     a_n = 2 + \sum_{i=1}^{n-1} 4^i = 2 + 4 \left(\frac{4^{n-1} - 1}{4 - 1}\right) = 2 + \frac{4^n - 4}{3}
     \]
     \[
     a_n = \frac{4^n - 4 + 6}{3} = \frac{4^n + 2}{3}
     \]

The final answer is $\boxed{\frac{4^n + 2}{3}}$.