Find all natural numbers $n$ for which there is a permutation $(p_1,p_2,...,p_n)$ of numbers $(1,2,...,n)$ such that sets $\{p_1 +1, p_2 + 2,..., p_n +n\}$ and $\{p_1-1, p_2-2,...,p_n -n\}$ are complete residue systems $\mod n$.
1. **Identify the problem and the given conditions:**
   We need to find all natural numbers \( n \) for which there exists a permutation \( (p_1, p_2, \ldots, p_n) \) of the numbers \( (1, 2, \ldots, n) \) such that the sets \( \{p_1 + 1, p_2 + 2, \ldots, p_n + n\} \) and \( \{p_1 - 1, p_2 - 2, \ldots, p_n - n\} \) are complete residue systems modulo \( n \).

2. **Analyze the conditions for the sets to be complete residue systems modulo \( n \):**
   - A complete residue system modulo \( n \) contains all integers from \( 0 \) to \( n-1 \) modulo \( n \).
   - Therefore, the sums \( p_i + i \) and \( p_i - i \) must cover all residues \( 0, 1, 2, \ldots, n-1 \) modulo \( n \).

3. **Sum of the elements in the sets:**
   - Consider the sum of the elements in the set \( \{p_1 + 1, p_2 + 2, \ldots, p_n + n\} \):
     \[
     \sum_{i=1}^n (p_i + i) = \sum_{i=1}^n p_i + \sum_{i=1}^n i
     \]
     Since \( (p_1, p_2, \ldots, p_n) \) is a permutation of \( (1, 2, \ldots, n) \), we have:
     \[
     \sum_{i=1}^n p_i = \sum_{i=1}^n i = \frac{n(n+1)}{2}
     \]
     Therefore:
     \[
     \sum_{i=1}^n (p_i + i) = \frac{n(n+1)}{2} + \frac{n(n+1)}{2} = n(n+1)
     \]
     Since the set \( \{p_1 + 1, p_2 + 2, \ldots, p_n + n\} \) is a complete residue system modulo \( n \), the sum of its elements modulo \( n \) must be congruent to the sum of the first \( n \) natural numbers modulo \( n \):
     \[
     n(n+1) \equiv 0 \pmod{n}
     \]
     This is always true since \( n(n+1) \) is divisible by \( n \).

4. **Sum of the squares of the elements in the sets:**
   - Consider the sum of the squares of the elements in the set \( \{p_1 + 1, p_2 + 2, \ldots, p_n + n\} \):
     \[
     \sum_{i=1}^n (p_i + i)^2
     \]
     Similarly, for the set \( \{p_1 - 1, p_2 - 2, \ldots, p_n - n\} \):
     \[
     \sum_{i=1}^n (p_i - i)^2
     \]
     Adding these two sums, we get:
     \[
     \sum_{i=1}^n (p_i + i)^2 + \sum_{i=1}^n (p_i - i)^2
     \]
     Expanding and simplifying:
     \[
     \sum_{i=1}^n (p_i^2 + 2p_i i + i^2) + \sum_{i=1}^n (p_i^2 - 2p_i i + i^2) = 2 \sum_{i=1}^n p_i^2 + 2 \sum_{i=1}^n i^2
     \]
     Since \( \sum_{i=1}^n p_i^2 = \sum_{i=1}^n i^2 \) for a permutation:
     \[
     2 \sum_{i=1}^n i^2 + 2 \sum_{i=1}^n i^2 = 4 \sum_{i=1}^n i^2
     \]
     The sum of the squares of the first \( n \) natural numbers is:
     \[
     \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}
     \]
     Therefore:
     \[
     4 \sum_{i=1}^n i^2 = 4 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3}
     \]
     For this to be congruent to \( 0 \pmod{n} \), \( n \) must not be divisible by \( 2 \) or \( 3 \).

5. **Conclusion:**
   - The conditions \( 2 \nmid n \) and \( 3 \nmid n \) imply that \( n \) must be of the form \( 6k \pm 1 \).
   - Therefore, the natural numbers \( n \) for which there exists such a permutation are those that satisfy \( n \equiv 1 \) or \( n \equiv 5 \pmod{6} \).

The final answer is \( \boxed{ n \equiv 1 } \) or \( n \equiv 5 \pmod{6} \).