In a bag are all natural numbers less than or equal to $999$ whose digits sum to $6$. What is the probability of drawing a number from the bag that is divisible by $11$?
1. Let the number be $\overline{abc}$, where $a$, $b$, and $c$ are digits. We are given that the sum of the digits is $6$, i.e., 
   \[
   a + b + c = 6
   \]
   Additionally, for the number to be divisible by $11$, the alternating sum of its digits must be a multiple of $11$. This gives us the equation:
   \[
   a - b + c = m
   \]
   where $m$ is some multiple of $11$.

2. Adding the two equations together, we get:
   \[
   (a + b + c) + (a - b + c) = 6 + m
   \]
   Simplifying, we obtain:
   \[
   2a + 2c = 6 + m
   \]
   Dividing by $2$, we get:
   \[
   a + c = \frac{6 + m}{2}
   \]
   For $a + c$ to be an integer, $6 + m$ must be even, implying that $m$ must be even. Since $m$ is a multiple of $11$, the only even multiple of $11$ is $0$ (as $22$ and $-22$ would make $a + c$ exceed the digit range).

3. Setting $m = 0$, we get:
   \[
   a + c = \frac{6 + 0}{2} = 3
   \]
   Substituting back into the original sum equation, we get:
   \[
   b = 6 - (a + c) = 6 - 3 = 3
   \]
   So, $b = 3$. We now need to find the number of nonnegative integer solutions to $a + c = 3$. The possible pairs $(a, c)$ are:
   \[
   (0, 3), (1, 2), (2, 1), (3, 0)
   \]
   Thus, there are $4$ solutions.

4. Next, we need to find the total number of possibilities for $a + b + c = 6$. Using the stars and bars method, the number of nonnegative integer solutions is given by:
   \[
   \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} = \frac{8 \times 7}{2} = 28
   \]

5. Finally, the probability of drawing a number from the bag that is divisible by $11$ is:
   \[
   \frac{4}{28} = \frac{1}{7}
   \]

The final answer is $\boxed{\frac{1}{7}}$