A natural number $n$ is at least two digits long. If we write a certain digit between the tens digit and the units digit of this number, we obtain six times the number $n$. Find all numbers $n$ with this property.
1. Let \( n \) be a two-digit number, which can be expressed as \( n = 10a + b \), where \( a \) and \( b \) are digits, i.e., \( 1 \leq a \leq 9 \) and \( 0 \leq b \leq 9 \).

2. According to the problem, if we insert a certain digit \( c \) between the tens digit \( a \) and the units digit \( b \) of \( n \), we obtain a new number. This new number can be written as \( 100a + 10c + b \).

3. The problem states that this new number is six times the original number \( n \). Therefore, we have the equation:
   \[
   100a + 10c + b = 6(10a + b)
   \]

4. Simplify the equation:
   \[
   100a + 10c + b = 60a + 6b
   \]
   \[
   100a + 10c + b - 60a - 6b = 0
   \]
   \[
   40a + 10c - 5b = 0
   \]
   \[
   8a + 2c - b = 0
   \]
   \[
   b = 8a + 2c
   \]

5. Since \( b \) is a digit, \( 0 \leq b \leq 9 \). Therefore, we need to find values of \( a \) and \( c \) such that \( 8a + 2c \) is a digit.

6. Let's test possible values of \( a \) and \( c \):
   - For \( a = 1 \):
     \[
     b = 8(1) + 2c = 8 + 2c
     \]
     \[
     0 \leq 8 + 2c \leq 9
     \]
     \[
     -8 \leq 2c \leq 1
     \]
     \[
     -4 \leq c \leq 0.5
     \]
     Since \( c \) must be a digit, the only possible value is \( c = 0 \):
     \[
     b = 8 + 2(0) = 8
     \]
     Therefore, \( n = 10(1) + 8 = 18 \).

   - For \( a = 2 \):
     \[
     b = 8(2) + 2c = 16 + 2c
     \]
     \[
     0 \leq 16 + 2c \leq 9
     \]
     \[
     -16 \leq 2c \leq -7
     \]
     This range is not possible for \( c \) since \( c \) must be a digit.

   - For \( a \geq 3 \):
     \[
     b = 8a + 2c
     \]
     \[
     8a + 2c \geq 24
     \]
     This is not possible since \( b \) must be a digit (i.e., \( 0 \leq b \leq 9 \)).

7. Therefore, the only solution is \( n = 18 \).

The final answer is \( \boxed{18} \)