Suppose that the mean and median of the non-negative numbers 21, 8,...

The given numbers are $21, 8, 17, a, 51, 103, b, 13, 67$. There are $9$ numbers in total.

Given that the mean is $40$, the sum of the numbers is:
$\sum x_i = 9 \times 40 = 360$

The sum of the known numbers is:
$21 + 8 + 17 + 51 + 103 + 13 + 67 = 280$

Therefore, the sum of $a$ and $b$ is:
$a + b = 360 - 280 = 80$

The median of the $9$ numbers is $21$. Arranging the known numbers in ascending order, we get:
$8, 13, 17, 21, 51, 67, 103$

There are $3$ numbers less than $21$ and $3$ numbers greater than $21$. For $21$ to be the median (the $5^{\text{th}}$ observation), one of the unknown numbers must be $\le 21$ and the other must be $\ge 21$. Since $a > b$, we must have:
$a \ge 21$ and $b \le 21$

The mean deviation about the median is $26$, so the sum of absolute deviations from the median is:
$\sum |x_i - 21| = 9 \times 26 = 234$

The sum of absolute deviations of the known numbers from $21$ is:
$|8 - 21| + |13 - 21| + |17 - 21| + |21 - 21| + |51 - 21| + |67 - 21| + |103 - 21|$
$= 13 + 8 + 4 + 0 + 30 + 46 + 82 = 183$

The sum of absolute deviations of $a$ and $b$ from $21$ is:
$|a - 21| + |b - 21| = 234 - 183 = 51$

Since $a \ge 21$ and $b \le 21$, we can remove the absolute values:
$(a - 21) + (21 - b) = 51$
$a - b = 51$

We now have a system of two equations:
$a + b = 80$
$a - b = 51$

Adding the two equations, we get:
$2a = 131$

Answer: $131$