Let $n$ be a positive integer not divisible by $3$. A triangular grid of length $n$ is obtained by dissecting a regular triangle with length $n$ into $n^2$ unit regular triangles. There is an orange at each vertex of the grid, which sums up to 
\[\frac{(n+1)(n+2)}{2}\]
oranges. A triple of oranges $A,B,C$ is [i]good[/i] if each $AB,AC$ is some side of some unit regular triangles, and $\angle BAC = 120^{\circ}$. Each time, Yen can take away a good triple of oranges from the grid. Determine the maximum number of oranges Yen can take.
1. **Understanding the Problem:**
   - We have a triangular grid of side length \( n \), dissected into \( n^2 \) unit regular triangles.
   - There is an orange at each vertex of the grid, totaling \(\frac{(n+1)(n+2)}{2}\) oranges.
   - A triple of oranges \( A, B, C \) is considered *good* if \( AB \) and \( AC \) are sides of some unit regular triangles and \(\angle BAC = 120^\circ\).
   - Yen can take away a good triple of oranges each time.
   - We need to determine the maximum number of oranges Yen can take.

2. **Total Number of Oranges:**
   - The total number of vertices (oranges) in the grid is given by:
     \[
     \frac{(n+1)(n+2)}{2}
     \]

3. **Divisibility by 3:**
   - Since Yen can only take away oranges in groups of 3, the number of oranges he can take must be divisible by 3.
   - The total number of oranges \(\frac{(n+1)(n+2)}{2}\) must also be divisible by 3.

4. **Upper Bound:**
   - We claim that the maximum number of oranges Yen can take is \(\min\{\alpha, \frac{(n+1)(n+2)}{2} - 3\}\), where \(\alpha\) denotes Yen's maximal capacity.
   - The bound \(\alpha\) is true by definition, and the construction in all situations is direct.
   - It suffices to prove an upper bound of \(\frac{(n+1)(n+2)}{2} - 3\) assuming \(\alpha \ge \frac{(n+1)(n+2)}{2} - 3\).

5. **Proof by Contradiction:**
   - Suppose Yen can take all oranges on all 3 sides.
   - Consider adding the \(120^\circ\) angles that cover the bottommost line one by one from left to right.
   - Label the points \(1, 2, \ldots, n+1\) in order.
   - In each step, add the angle covering the point with the smallest label not yet covered.
   - We claim that each time after one or two new angles are added, the point having the largest label among those already covered satisfies the following property: the point immediately to the right and above it is also covered.
   - Starting from one such stage (including the initial stage), there are 2 possible ways to put in the angle we want to add.
   - If it does not cover the 2nd-largest-label-not-yet-covered, then there is only one way to add in a second angle; otherwise, we are already done.
   - In either case, our claim is true, which brings a contradiction if all labels are eventually covered.

6. **Conclusion:**
   - Therefore, Yen cannot take all oranges, and the maximum number of oranges he can take is \(\frac{(n+1)(n+2)}{2} - 3\).

The final answer is \(\boxed{\frac{(n+1)(n+2)}{2} - 3}\).