A $5\times5$ grid of squares is filled with integers. Call a rectangle [i]corner-odd[/i] if its sides are grid lines and the sum of the integers in its four corners is an odd number. What is the maximum possible number of corner-odd rectangles within the grid?

Note: A rectangles must have four distinct corners to be considered [i]corner-odd[/i]; i.e. no $1\times k$ rectangle can be [i]corner-odd[/i] for any positive integer $k$.
1. **Understanding the Problem:**
   We need to find the maximum number of corner-odd rectangles in a $5 \times 5$ grid. A rectangle is corner-odd if the sum of the integers at its four corners is odd.

2. **Analyzing the Grid:**
   Consider a $5 \times 5$ grid. Each cell can contain either an odd or an even integer. For a rectangle to be corner-odd, the sum of the integers at its four corners must be odd. This implies that the number of odd integers among the four corners must be odd (either 1 or 3).

3. **Pairing Rows and Columns:**
   Let's take a pair of rows and consider the sums of the two numbers in each of the 5 columns. Each sum can be either odd or even. We want to maximize the number of odd sums.

4. **Maximizing Odd Sums:**
   For each pair of rows, we can have a maximum of 2 odd sums and 3 even sums, or vice versa. The maximum number of odd sums occurs when we have 2 odd sums and 3 even sums, or 3 odd sums and 2 even sums. This is because the product $a(5-a)$, where $a$ is the number of odd sums, is maximized when $a = 2$ or $a = 3$.

5. **Calculating the Maximum Number of Corner-Odd Rectangles:**
   There are $\binom{5}{2} = 10$ ways to choose 2 rows from 5. For each pair of rows, we can have up to 6 corner-odd rectangles (since $2 \cdot 3 = 6$). Therefore, the maximum number of corner-odd rectangles is:
   \[
   10 \times 6 = 60
   \]

6. **Constructing the Grid:**
   To achieve this maximum, we can place the odd integers in such a way that maximizes the number of odd sums for each pair of rows. One way to do this is to place the odd integers along a long diagonal of the grid. This ensures that for each pair of rows, we have the maximum number of odd sums.

The final answer is $\boxed{60}$.