A $5$-digit number (in base $10$) has digits $k, k + 1, k + 2, 3k, k + 3$ in that order, from left to right. If this number is $m^2$ for some natural number $m$, find the sum of the digits of $m$.
1. We start by noting that the number is a 5-digit number with digits \( k, k+1, k+2, 3k, k+3 \) in that order. Therefore, the number can be written as:
   \[
   N = 10000k + 1000(k+1) + 100(k+2) + 10(3k) + (k+3)
   \]
   Simplifying this expression, we get:
   \[
   N = 10000k + 1000k + 1000 + 100k + 200 + 30k + k + 3 = 11131k + 1203
   \]

2. Since \( N \) is a 5-digit number, we need to ensure that \( 10000 \leq N < 100000 \). This gives us the inequality:
   \[
   10000 \leq 11131k + 1203 < 100000
   \]

3. Solving the lower bound:
   \[
   10000 \leq 11131k + 1203
   \]
   \[
   8797 \leq 11131k
   \]
   \[
   k \geq \frac{8797}{11131} \approx 0.79
   \]
   Since \( k \) is an integer, \( k \geq 1 \).

4. Solving the upper bound:
   \[
   11131k + 1203 < 100000
   \]
   \[
   11131k < 98797
   \]
   \[
   k < \frac{98797}{11131} \approx 8.88
   \]
   Since \( k \) is an integer, \( k \leq 8 \).

5. We also need to ensure that \( 3k \) is a single digit, which implies:
   \[
   3k \leq 9 \implies k \leq 3
   \]

6. Combining the constraints, we have \( 1 \leq k \leq 3 \). We will check each value of \( k \) to see if \( N \) is a perfect square.

   - For \( k = 1 \):
     \[
     N = 11131(1) + 1203 = 12334 \quad (\text{not a perfect square})
     \]

   - For \( k = 2 \):
     \[
     N = 11131(2) + 1203 = 23465 \quad (\text{not a perfect square})
     \]

   - For \( k = 3 \):
     \[
     N = 11131(3) + 1203 = 34596
     \]
     We need to check if 34596 is a perfect square:
     \[
     \sqrt{34596} = 186 \quad (\text{since } 186^2 = 34596)
     \]

7. Since \( 34596 \) is indeed a perfect square, we have \( m = 186 \).

8. Finally, we find the sum of the digits of \( m \):
   \[
   1 + 8 + 6 = 15
   \]

The final answer is \(\boxed{15}\).