Given is a positive integer $k$. There are $n$ points chosen on a line, such the distance between any two adjacent points is the same. The points are colored in $k$ colors. For each pair of monochromatic points such that there are no points of the same color between them, we record the distance between these two points. If all distances are distinct, find the largest possible $n$.
To solve this problem, we need to determine the largest possible number of points \( n \) on a line such that the distances between any two adjacent points are the same, and the points are colored in \( k \) colors. The distances between pairs of monochromatic points (points of the same color) must all be distinct.

1. **Construction**:
   We can construct a sequence of points colored in \( k \) colors as follows:
   \[
   1, 2, 3, \ldots, k, k, \ldots, 3, 2, 1, 2, 3, \ldots, k
   \]
   This sequence ensures that each color appears in a symmetric pattern, and the distances between monochromatic points are distinct.

2. **Bounding the Total Gap Length**:
   Let \( n \) be the total number of points. There are \( n - k \) gaps between the points. The sum of the first \( n - k \) natural numbers is:
   \[
   1 + 2 + \cdots + (n - k) = \frac{(n - k)(n - k + 1)}{2}
   \]
   This represents the minimum total gap length required.

3. **Calculating the Total Gap Length**:
   Let \( x_i \) be the position of the first occurrence of color \( i \), and \( y_i \) be the position of the last occurrence of color \( i \). The total gap length can be expressed as:
   \[
   \sum_{i=1}^k (y_i - x_i)
   \]
   Since the positions \( x_i \) and \( y_i \) are within the range of \( 1 \) to \( n \), the total gap length is bounded by:
   \[
   \sum_{i=1}^k y_i - \sum_{i=1}^k x_i \leq (n + (n-1) + \cdots + (n-k+1)) - (1 + 2 + \cdots + k)
   \]
   Simplifying the sums, we get:
   \[
   \sum_{i=1}^k y_i - \sum_{i=1}^k x_i \leq k(n - k)
   \]

4. **Combining the Inequalities**:
   We now combine the inequalities to find the upper bound for \( n \):
   \[
   k(n - k) \geq \frac{(n - k)(n - k + 1)}{2}
   \]
   Simplifying this inequality:
   \[
   2k(n - k) \geq (n - k)(n - k + 1)
   \]
   Dividing both sides by \( n - k \) (assuming \( n \neq k \)):
   \[
   2k \geq n - k + 1
   \]
   Rearranging terms, we get:
   \[
   n \leq 3k - 1
   \]

Thus, the largest possible \( n \) is \( 3k - 1 \).

The final answer is \(\boxed{3k - 1}\).