Let's solve this problem step by step.

Step 1: Understand the Sum and Average of Natural Numbers from 1 to $n$

The sum of the first $n$ natural numbers is given by: $S_n = \frac{n(n+1)}{2}$ The average of these $n$ numbers would be: $\text{Average} = \frac{S_n}{n} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}$

Step 2: Set up the Equation

Let’s denote the missed number by $m$. The boy calculated the sum without including $m$, so the sum he actually computed was: $S_n - m$ Since he averaged the remaining $n - 1$ numbers and got $33 \frac{9}{13}$ (or $\frac{438}{13}$), we set up the equation: $\frac{S_n - m}{n - 1} = \frac{438}{13}$

Step 3: Substitute $S_n = \frac{n(n+1)}{2}$

Substitute the expression for $S_n$: $\frac{\frac{n(n+1)}{2} - m}{n - 1} = \frac{438}{13}$

Step 4: Clear the Denominator by Cross Multiplying

Multiply both sides by $13(n - 1)$ to eliminate the fraction: $13 \left(\frac{n(n+1)}{2} - m\right) = 438(n - 1)$

Step 5: Solve for $n$ and $m$ Based on Additional Details

Given that the average without $m$ is close to 33, we infer that $n$ would likely be close to 66