Let $ n$ be a positive integer. Determine the number of triangles (non congruent) with integral side lengths and the longest side length is $ n$.
1. Let \( n \) be a positive integer. We need to determine the number of non-congruent triangles with integral side lengths where the longest side length is \( n \).

2. Consider a triangle with sides \( (x, y, n) \) where \( x \leq y \leq n \). By the triangle inequality, we have:
   \[
   x + y > n
   \]
   Since \( x \leq y \leq n \), we also have:
   \[
   x + y \geq n
   \]
   and
   \[
   x, y \leq n
   \]

3. Given \( y \), we need to find the range of \( x \). From the inequality \( x + y \geq n \), we get:
   \[
   x \geq n - y
   \]
   Since \( x \leq y \), we have:
   \[
   n - y \leq x \leq y
   \]

4. The number of possible values for \( x \) given \( y \) is:
   \[
   y - (n - y) + 1 = 2y - n + 1
   \]

5. Summing over all possible values of \( y \) from \( \lceil \frac{n}{2} \rceil \) to \( n \), we get:
   \[
   \sum_{y=\lceil \frac{n}{2} \rceil}^{n} (2y - n + 1)
   \]

6. We can split the sum into two parts:
   \[
   \sum_{y=\lceil \frac{n}{2} \rceil}^{n} 2y - \sum_{y=\lceil \frac{n}{2} \rceil}^{n} n + \sum_{y=\lceil \frac{n}{2} \rceil}^{n} 1
   \]

7. The first sum is:
   \[
   2 \sum_{y=\lceil \frac{n}{2} \rceil}^{n} y = 2 \left( \frac{n(n+1)}{2} - \frac{(\lceil \frac{n}{2} \rceil - 1) \lceil \frac{n}{2} \rceil}{2} \right)
   \]

8. The second sum is:
   \[
   n \left( n - \lceil \frac{n}{2} \rceil + 1 \right)
   \]

9. The third sum is:
   \[
   n - \lceil \frac{n}{2} \rceil + 1
   \]

10. Combining these, we get:
    \[
    2 \left( \frac{n(n+1)}{2} - \frac{(\lceil \frac{n}{2} \rceil - 1) \lceil \frac{n}{2} \rceil}{2} \right) - n \left( n - \lceil \frac{n}{2} \rceil + 1 \right) + n - \lceil \frac{n}{2} \rceil + 1
    \]

11. Simplifying, we get:
    \[
    \frac{(n+2)^2 - 2 \left\lfloor \frac{n}{2} \right\rfloor}{4}
    \]

The final answer is \(\boxed{\frac{(n+2)^2 - 2 \left\lfloor \frac{n}{2} \right\rfloor}{4}}\).