We call a permutation $ \left(a_1, a_2, ..., a_n\right)$ of $ \left(1, 2, ..., n\right)$ [i]quadratic[/i] if there exists at least a perfect square among the numbers $ a_1$, $ a_1 \plus{} a_2$, $ ...$, $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n$. Find all natural numbers $ n$ such that all permutations in $ S_n$ are quadratic.

[i]Remark.[/i] $ S_{n}$ denotes the $ n$-th symmetric group, the group of permutations on $ n$ elements.
To solve the problem, we need to determine for which natural numbers \( n \) every permutation of \( \{1, 2, \ldots, n\} \) contains at least one partial sum that is a perfect square. 

1. **Sum of First \( n \) Natural Numbers**:
   The sum of the first \( n \) natural numbers is given by:
   \[
   S_n = \frac{n(n+1)}{2}
   \]
   If \( S_n \) is a perfect square, then there exists an integer \( k \) such that:
   \[
   \frac{n(n+1)}{2} = k^2
   \]

2. **Perfect Square Condition**:
   We need to find \( n \) such that \( \frac{n(n+1)}{2} \) is a perfect square. This can be rewritten as:
   \[
   n(n+1) = 2k^2
   \]
   This is a Diophantine equation, and we need to find integer solutions for \( n \) and \( k \).

3. **Pell's Equation**:
   The equation \( n(n+1) = 2k^2 \) can be transformed into a Pell's equation. By setting \( x = 2n+1 \) and \( y = 2k \), we get:
   \[
   x^2 - 2y^2 = 1
   \]
   This is a standard Pell's equation, and its solutions can be generated using the fundamental solution \( (x_1, y_1) = (3, 2) \) and the recurrence relations:
   \[
   x_{n+1} = 3x_n + 4y_n
   \]
   \[
   y_{n+1} = 2x_n + 3y_n
   \]

4. **Generating Solutions**:
   Starting with the fundamental solution \( (x_1, y_1) = (3, 2) \), we can generate further solutions. For example:
   \[
   (x_2, y_2) = (17, 12)
   \]
   \[
   (x_3, y_3) = (99, 70)
   \]
   These correspond to:
   \[
   n_1 = \frac{3-1}{2} = 1
   \]
   \[
   n_2 = \frac{17-1}{2} = 8
   \]
   \[
   n_3 = \frac{99-1}{2} = 49
   \]
   Thus, \( n = 1, 8, 49, \ldots \) are solutions where \( \frac{n(n+1)}{2} \) is a perfect square.

5. **Conclusion**:
   For these values of \( n \), every permutation of \( \{1, 2, \ldots, n\} \) will contain at least one partial sum that is a perfect square. This is because the sum of all elements is a perfect square, and any permutation will have at least one partial sum that is a perfect square.

The final answer is \( \boxed{ n = 1, 8, 49, \ldots } \)