To solve this problem, we need to minimize the sum of the products of the entries in the $1 \times 2$ rectangles. Let's break down the steps to achieve this:

1. Define the Matrix:
The matrix $\{a_{ij}\}_{i,j=0}^{9}$ is given by $a_{ij} = 10i + j + 1$. This means each element in the matrix can be calculated using this formula.

2. Calculate the Products:
For each $1 \times 2$ rectangle, we need to calculate the product of the two adjacent entries. There are two possible orientations for each rectangle:
- Horizontal: covering entries $a_{ij}$ and $a_{i(j+1)}$
- Vertical: covering entries $a_{ij}$ and $a_{(i+1)j}$

3. Compare the Products:
We need to compare the sum of the products for different configurations to find the minimal possible sum. Let's consider the products for both horizontal and vertical rectangles:
- Horizontal product: $a_{ij} \cdot a_{i(j+1)}$
- Vertical product: $a_{ij} \cdot a_{(i+1)j}$

4. Observation and Comparison:
We need to compare the sum of products for different configurations. Let's consider the following observation:
$$a_{ij} \cdot a_{i(j+1)} + a_{(i+1)j} \cdot a_{(i+1)(j+1)} > a_{ij} \cdot a_{(i+1)j} + a_{i(j+1)} \cdot a_{(i+1)(j+1)}$$
This inequality suggests that the sum of products is minimized when we use vertical rectangles rather than horizontal rectangles.

5. Summing the Products:
To minimize the sum of the products, we should use vertical rectangles. Let's calculate the sum of the products for vertical rectangles:
$$\sum_{i=0}^{8} \sum_{j=0}^{9} a_{ij} \cdot a_{(i+1)j}$$
Each vertical rectangle covers two adjacent entries in the same column.

6. Calculate the Sum:
We need to calculate the sum of the products for all vertical rectangles:
$$\sum_{i=0}^{8} \sum_{j=0}^{9} (10i + j + 1) \cdot (10(i+1) + j + 1)$$
Simplifying the expression inside the sum:
$$(10i + j + 1) \cdot (10(i+1) + j + 1) = (10i + j + 1) \cdot (10i + 10 + j + 1) = (10i + j + 1) \cdot (10i + j + 11)$$
Expanding the product:
$$(10i + j + 1)(10i + j + 11) = 100i^2 + 20ij + 10i + 10ij + j^2 + 11j + 10i + j + 11 = 100i^2 + 30ij + j^2 + 21j + 20i + 11$$

7. Summing Over All Rectangles:
We need to sum this expression over all valid $i$ and $j$:
$$\sum_{i=0}^{8} \sum_{j=0}^{9} (100i^2 + 30ij + j^2 + 21j + 20i + 11)$$

8. Final Calculation:
This double sum can be computed by summing each term separately:
$$\sum_{i=0}^{8} \sum_{j=0}^{9} 100i^2 + \sum_{i=0}^{8} \sum_{j=0}^{9} 30ij + \sum_{i=0}^{8} \sum_{j=0}^{9} j^2 + \sum_{i=0}^{8} \sum_{j=0}^{9} 21j + \sum_{i=0}^{8} \sum_{j=0}^{9} 20i + \sum_{i=0}^{8} \sum_{j=0}^{9} 11$$

Each of these sums can be computed separately, but the key observation is that using vertical rectangles minimizes the sum of the products.