For each positive integer $ n$,denote by $ S(n)$ the sum of all digits in decimal representation of $ n$.
Find all positive integers $ n$,such that $ n\equal{}2S(n)^3\plus{}8$.
1. **Establishing the inequality:**
   Let \( d \) be the number of digits in \( n \). The sum of the digits \( S(n) \) is at most \( 9d \) because each digit can be at most 9. Therefore, we have:
   \[
   S(n) \leq 9d
   \]
   On the other hand, the smallest \( n \) with \( d \) digits is \( 10^{d-1} \). Thus, we have:
   \[
   n \geq 10^{d-1}
   \]
   Given \( n = 2S(n)^3 + 8 \), we substitute \( S(n) \leq 9d \) into the equation:
   \[
   n \leq 2(9d)^3 + 8 = 2 \cdot 729d^3 + 8 = 1458d^3 + 8
   \]
   Combining the inequalities, we get:
   \[
   10^{d-1} \leq 1458d^3 + 8
   \]
   We need to find the maximum \( d \) for which this inequality holds.

2. **Checking the inequality for \( d \geq 8 \):**
   For \( d = 8 \):
   \[
   10^{8-1} = 10^7 = 10000000
   \]
   \[
   1458 \cdot 8^3 + 8 = 1458 \cdot 512 + 8 = 746496 + 8 = 746504
   \]
   Clearly, \( 10000000 > 746504 \), so the inequality does not hold for \( d \geq 8 \). Therefore, \( d \) must be less than 8, implying \( S(n) \leq 9 \times 7 = 63 \).

3. **Considering the congruence modulo 9:**
   Since \( S(n) \equiv n \pmod{9} \), we have:
   \[
   n \equiv 2S(n)^3 + 8 \pmod{9}
   \]
   We need to solve:
   \[
   n \equiv 2n^3 + 8 \pmod{9}
   \]
   Testing values modulo 9, we find that the only solution is:
   \[
   n \equiv 1 \pmod{9}
   \]
   Therefore, \( S(n) \) must be one of the values that make \( n \equiv 1 \pmod{9} \). The possible values of \( S(n) \) are 1, 10, 19, 28, 37, 46, 55.

4. **Checking the possible values of \( S(n) \):**
   - For \( S(n) = 1 \):
     \[
     n = 2 \cdot 1^3 + 8 = 2 \cdot 1 + 8 = 10
     \]
     \( S(10) = 1 \), so \( n = 10 \) is a solution.
   - For \( S(n) = 10 \):
     \[
     n = 2 \cdot 10^3 + 8 = 2 \cdot 1000 + 8 = 2008
     \]
     \( S(2008) = 2 + 0 + 0 + 8 = 10 \), so \( n = 2008 \) is a solution.
   - For \( S(n) = 19 \):
     \[
     n = 2 \cdot 19^3 + 8 = 2 \cdot 6859 + 8 = 13726
     \]
     \( S(13726) = 1 + 3 + 7 + 2 + 6 = 19 \), so \( n = 13726 \) is a solution.
   - For \( S(n) = 28, 37, 46, 55 \):
     These values are not possible because \( S(n) \leq 63 \) and \( n \equiv 1 \pmod{9} \). Additionally, \( 55 > 9 \times 6 \).

The final answer is \( \boxed{ n = 10, 2008, 13726 } \).