We choose 100 points in the coordinate plane. Let $N$ be the number of triples $(A,B,C)$ of distinct chosen points such that $A$ and $B$ have the same $y$-coordinate, and $B$ and $C$ have the same $x$-coordinate. Find the greatest value that $N$ can attain considering all possible ways to choose the points.
1. **Claim and Initial Setup**: We claim that the maximum value of \( N \) is \( \boxed{8100} \). To see this, consider a \( 10 \times 10 \) grid. Each rectangle of four points contributes exactly 4 such triples (the triples correspond to right triangles with legs parallel to the axes). There are \(\binom{10}{2}^2\) rectangles, so there are \( 4 \binom{10}{2}^2 = 8100 \) right triangles.

2. **General Case Analysis**: Let \( y_1, \ldots, y_r \) be the distinct \( y \)-coordinates in all the 100 points, and let \( m_i \) be the number of points with \( y \)-coordinate equal to \( y_i \). Focusing on some particular \( i \), we have \( m_i \) points all on a horizontal line. Let \( n_j \) for \( 1 \le j \le m_i \) be the number of points on the perpendicular line passing through the \( j \)-th point not equal to the point on the horizontal line. We see that the number of triples with \( y \)-coordinate equal to \( y_i \) is:
   \[
   n_1(m_i-1) + \cdots + n_{m_i}(m_i-1) = (m_i-1)(n_1 + \cdots + n_{m_i}).
   \]

3. **Bounding the Sum**: We have that \( n_1 + \cdots + n_{m_i} \) is at most \( 100 - m_i \), so the total number of points is at most:
   \[
   N \le \sum_{i=1}^r (m_i-1)(100 - m_i).
   \]

4. **Concavity and Jensen's Inequality**: Note that \( f(m) = (m-1)(100-m) \) is a concave function as \( f''(m) = -1 \). By Jensen's inequality, we have:
   \[
   N \le r \left( \frac{100}{r} - 1 \right) \left( 100 - \frac{100}{r} \right).
   \]
   Since \( \sum m_i = 100 \), we can simplify:
   \[
   N \le 100 \left( \frac{100}{r} - 1 \right) (r - 1).
   \]

5. **Maximizing the Expression**: Note that:
   \[
   \left( \frac{100}{r} - 1 \right) (r - 1) = 101 - \left( \frac{100}{r} + r \right).
   \]
   By the AM-GM inequality, we have:
   \[
   \frac{100}{r} + r \ge 2 \sqrt{100} = 20.
   \]
   Therefore:
   \[
   101 - \left( \frac{100}{r} + r \right) \le 101 - 20 = 81.
   \]
   So:
   \[
   N \le 100 \cdot 81 = 8100.
   \]

This completes the proof.