For all positive integer $n$, define $f_n(x)$ such that $f_n(x) = \sum_{k=1}^n{|x - k|}$.
Determine all solution from the inequality $f_n(x) < 41$ for all positive $2$-digit integers $n$ (in decimal notation).
To solve the problem, we need to analyze the function \( f_n(x) = \sum_{k=1}^n |x - k| \) and determine the values of \( x \) for which \( f_n(x) < 41 \) for all positive 2-digit integers \( n \).

1. **Understanding the Function \( f_n(x) \)**:
   - The function \( f_n(x) \) is a sum of absolute values, which is piecewise linear with changes in slope at each integer \( k \) from 1 to \( n \).
   - The function is symmetric about \( x = \frac{n+1}{2} \).

2. **Case 1: \( n = 2m + 1 \) (Odd \( n \))**:
   - For odd \( n \), the minimum value of \( f_n(x) \) occurs at \( x = m + 1 \), where \( m = \frac{n-1}{2} \).
   - The minimum value is:
     \[
     f_{2m+1}(m+1) = \sum_{k=1}^{2m+1} |m+1 - k| = \sum_{k=1}^m (m+1 - k) + \sum_{k=m+2}^{2m+1} (k - m - 1)
     \]
     \[
     = m(m+1)
     \]
   - For \( f_{2m+1}(x) < 41 \), we need \( m(m+1) < 41 \). Solving for \( m \):
     \[
     m^2 + m - 41 < 0
     \]
     The roots of the quadratic equation \( m^2 + m - 41 = 0 \) are:
     \[
     m = \frac{-1 \pm \sqrt{1 + 4 \cdot 41}}{2} = \frac{-1 \pm \sqrt{165}}{2}
     \]
     Approximating \( \sqrt{165} \approx 12.845 \):
     \[
     m \approx \frac{-1 + 12.845}{2} \approx 5.922 \quad \text{and} \quad m \approx \frac{-1 - 12.845}{2} \approx -6.922
     \]
     Since \( m \) must be a positive integer, \( m \leq 5 \).

   - For \( m = 5 \), \( n = 2m + 1 = 11 \):
     \[
     f_{11}(x) = \sum_{k=1}^{11} |x - k|
     \]
     We need to find the range of \( x \) such that \( f_{11}(x) < 41 \).

     - For \( x \in [6, 7] \):
       \[
       f_{11}(x) = x + 24 \quad \text{(values in } [30, 31])
       \]
     - For \( x \in [7, 8] \):
       \[
       f_{11}(x) = 3x + 10 \quad \text{(values in } [31, 34])
       \]
     - For \( x \in [8, 9] \):
       \[
       f_{11}(x) = 5x - 6 \quad \text{(values in } [34, 39])
       \]
     - For \( x \in [9, 10] \):
       \[
       f_{11}(x) = 7x - 24 \quad \text{(values in } [39, 46])
       \]

     Therefore, \( x \) must satisfy:
     \[
     x < \frac{65}{7} \quad \text{and} \quad x > \frac{19}{7}
     \]
     \[
     x \in \left( \frac{19}{7}, \frac{65}{7} \right)
     \]

     Thus, the solution for \( n = 11 \) is:
     \[
     \boxed{\text{S1 : } n = 11 \text{ and } x \in \left( \frac{19}{7}, \frac{65}{7} \right)}
     \]

3. **Case 2: \( n = 2m \) (Even \( n \))**:
   - For even \( n \), the minimum value of \( f_n(x) \) occurs at \( x = m \) or \( x = m + 1 \), where \( m = \frac{n}{2} \).
   - The minimum value is:
     \[
     f_{2m}(m) = f_{2m}(m+1) = \sum_{k=1}^{2m} |m - k| = m^2
     \]
   - For \( f_{2m}(x) < 41 \), we need \( m^2 < 41 \). Solving for \( m \):
     \[
     m < \sqrt{41} \approx 6.4
     \]
     Since \( m \) must be a positive integer, \( m \leq 6 \).

   - For \( m = 5 \), \( n = 10 \):
     \[
     f_{10}(x) = \sum_{k=1}^{10} |x - k|
     \]
     We need to find the range of \( x \) such that \( f_{10}(x) < 41 \).

     - For \( x \in [5, 6] \):
       \[
       f_{10}(x) = 25
       \]
     - For \( x \in [6, 7] \):
       \[
       f_{10}(x) = 2x + 13 \quad \text{(values in } [25, 27])
       \]
     - For \( x \in [7, 8] \):
       \[
       f_{10}(x) = 4x - 1 \quad \text{(values in } [27, 31])
       \]
     - For \( x \in [8, 9] \):
       \[
       f_{10}(x) = 6x - 17 \quad \text{(values in } [31, 37])
       \]
     - For \( x \in [9, 10] \):
       \[
       f_{10}(x) = 8x - 35 \quad \text{(values in } [37, 45])
       \]

     Therefore, \( x \) must satisfy:
     \[
     x < \frac{19}{2} \quad \text{and} \quad x > \frac{3}{2}
     \]
     \[
     x \in \left( \frac{3}{2}, \frac{19}{2} \right)
     \]

     Thus, the solution for \( n = 10 \) is:
     \[
     \boxed{\text{S2 : } n = 10 \text{ and } x \in \left( \frac{3}{2}, \frac{19}{2} \right)}
     \]

   - For \( m = 6 \), \( n = 12 \):
     \[
     f_{12}(x) = \sum_{k=1}^{12} |x - k|
     \]
     We need to find the range of \( x \) such that \( f_{12}(x) < 41 \).

     - For \( x \in [6, 7] \):
       \[
       f_{12}(x) = 36
       \]
     - For \( x \in [7, 8] \):
       \[
       f_{12}(x) = 2x + 22 \quad \text{(values in } [36, 38])
       \]
     - For \( x \in [8, 9] \):
       \[
       f_{12}(x) = 4x + 6 \quad \text{(values in } [38, 42])
       \]

     Therefore, \( x \) must satisfy:
     \[
     x < \frac{35}{4} \quad \text{and} \quad x > \frac{17}{4}
     \]
     \[
     x \in \left( \frac{17}{4}, \frac{35}{4} \right)
     \]

     Thus, the solution for \( n = 12 \) is:
     \[
     \boxed{\text{S3 : } n = 12 \text{ and } x \in \left( \frac{17}{4}, \frac{35}{4} \right)}
     \]