We are given that six positive numbers have a mean of 12 and a range of 19. Additionally, four of the numbers are 13, 8, 16, and 4. We need to find the smallest of the other two numbers.

Step 1: Calculate the total sum of the six numbers

The mean of the six numbers is given as 12. The mean is the total sum divided by the number of elements. So, we can calculate the total sum:

$\text{Mean} = \frac{\text{Sum of the numbers}}{6}$ $12 = \frac{\text{Sum of the numbers}}{6}$ $\text{Sum of the numbers} = 12 \times 6 = 72$

Step 2: Calculate the sum of the four given numbers

The four given numbers are 13, 8, 16, and 4. Let's calculate their sum:

$\text{Sum of the four numbers} = 13 + 8 + 16 + 4 = 41$

Step 3: Calculate the sum of the remaining two numbers

We know that the sum of all six numbers is 72. The sum of the four given numbers is 41, so the sum of the remaining two numbers must be:

$\text{Sum of the remaining two numbers} = 72 - 41 = 31$

Step 4: Use the range to find the numbers

The range is the difference between the largest and smallest numbers. We are told the range is 19. The largest number among the given four is 16, so one of the remaining two numbers must be $16 + 19 = 35$, because the range is 19.

Now, the sum of the remaining two numbers is 31, so if one number is 35, the other number must be:

$\text{Other number} = 31 - 35 = -4$

Since a negative number is not possible for this problem, we must have made a mistake somewhere. We can try using the proper range at