How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$, inclusive?

$\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300$
1. **Calculate the total number of triangles:**
   We start by calculating the total number of ways to choose 3 points out of the 25 lattice points. This is given by the binomial coefficient:
   \[
   \binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300
   \]

2. **Subtract the number of collinear points:**
   We need to subtract the number of sets of 3 collinear points, as these do not form triangles with positive area.

   - **Horizontal and vertical lines:**
     Each row and each column has 5 points. The number of ways to choose 3 points from 5 is:
     \[
     \binom{5}{3} = 10
     \]
     Since there are 5 rows and 5 columns, the total number of collinear sets in rows and columns is:
     \[
     10 \times 5 + 10 \times 5 = 100
     \]

   - **Diagonals with slope 1 or -1:**
     We need to consider diagonals of the grid. Diagonals can be of different lengths, and we need to count the number of ways to choose 3 points from each diagonal.

     - **Diagonals with slope 1:**
       There are 9 diagonals with slope 1. The lengths of these diagonals are 1, 2, 3, 4, 5, 4, 3, 2, 1. Only diagonals of length 3 or more can have 3 collinear points.
       \[
       \binom{3}{3} = 1, \quad \binom{4}{3} = 4, \quad \binom{5}{3} = 10
       \]
       So, the total number of collinear sets in diagonals with slope 1 is:
       \[
       1 + 4 + 10 + 4 + 1 = 20
       \]

     - **Diagonals with slope -1:**
       Similarly, there are 9 diagonals with slope -1, and the count is the same as for slope 1:
       \[
       1 + 4 + 10 + 4 + 1 = 20
       \]

     Therefore, the total number of collinear sets in diagonals is:
     \[
     20 + 20 = 40
     \]

3. **Calculate the final number of triangles:**
   Subtract the number of collinear sets from the total number of triangles:
   \[
   2300 - 100 - 40 = 2160
   \]

The final answer is $\boxed{2160}$