A plane rectangular grid is given and a “rational point” is defined as a point $(x, y)$ where $x$ and $y$ are both rational numbers. Let $A,B,A',B'$ be four distinct rational points. Let $P$ be a point such that $\frac{A'B'}{AB}=\frac{B'P}{BP} = \frac{PA'}{PA}.$ In other words, the triangles $ABP, A'B'P$ are directly or oppositely similar. Prove that $P$ is in general a rational point and find the exceptional positions of $A'$ and $B'$ relative to $A$ and $B$ such that there exists a $P$ that is not a rational point.
1. **Given Information and Definitions:**
   - We have four distinct rational points \( A, B, A', B' \) on a plane rectangular grid.
   - A rational point is defined as a point \((x, y)\) where both \(x\) and \(y\) are rational numbers.
   - We need to prove that point \(P\) is generally a rational point and identify any exceptional positions of \(A'\) and \(B'\) relative to \(A\) and \(B\) where \(P\) is not a rational point.

2. **Similarity Condition:**
   - The triangles \(ABP\) and \(A'B'P\) are similar, meaning:
     \[
     \frac{A'B'}{AB} = \frac{B'P}{BP} = \frac{PA'}{PA}
     \]
   - This implies that the triangles are either directly similar (same orientation) or oppositely similar (one is a mirror image of the other).

3. **Coordinate Representation:**
   - Let \(A = (x_1, y_1)\), \(B = (x_2, y_2)\), \(A' = (x_1', y_1')\), and \(B' = (x_2', y_2')\).
   - Since \(A, B, A', B'\) are rational points, \(x_1, y_1, x_2, y_2, x_1', y_1', x_2', y_2'\) are all rational numbers.

4. **Finding Point \(P\):**
   - Let \(P = (x, y)\).
   - Using the similarity condition, we can set up the following equations based on the ratios of the distances:
     \[
     \frac{\sqrt{(x_1' - x_2')^2 + (y_1' - y_2')^2}}{\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}} = \frac{\sqrt{(x_2' - x)^2 + (y_2' - y)^2}}{\sqrt{(x_2 - x)^2 + (y_2 - y)^2}} = \frac{\sqrt{(x - x_1')^2 + (y - y_1')^2}}{\sqrt{(x - x_1)^2 + (y - y_1)^2}}
     \]

5. **Solving for \(P\):**
   - To solve for \(P\), we need to solve the system of equations derived from the similarity conditions. This involves equating the ratios and solving for \(x\) and \(y\).
   - Given that all coordinates involved are rational, the resulting equations will be in terms of rational numbers.

6. **Rationality of \(P\):**
   - Since the equations are derived from rational coordinates and involve only rational operations (addition, subtraction, multiplication, division), the solutions \(x\) and \(y\) will also be rational, provided the system of equations is consistent and solvable.

7. **Exceptional Cases:**
   - The exceptional cases occur when the system of equations is not solvable or leads to irrational solutions.
   - This can happen if \(A'\) and \(B'\) are positioned such that the similarity condition cannot be satisfied with rational coordinates for \(P\).
   - For example, if \(A'\) and \(B'\) are such that the ratio \(\frac{A'B'}{AB}\) is an irrational number, then \(P\) may not be a rational point.

Conclusion:
- Generally, \(P\) is a rational point because the equations derived from the similarity conditions involve only rational numbers.
- The exceptional positions of \(A'\) and \(B'\) relative to \(A\) and \(B\) occur when the ratio \(\frac{A'B'}{AB}\) is irrational, leading to an irrational solution for \(P\).

The final answer is \( \boxed{ P } \) is generally a rational point, except when \(\frac{A'B'}{AB}\) is irrational.