Problem 1678

1. Define the variables and the structure of the table:
Let $a^j_i$ denote the element in the $i$-th row and the $j$-th column of the rectangle. Let $n$ be the number of rows and $m$ be the number of columns. The total number of elements in the table is $n \times m = 221$. The sum of all elements in the table is $S = 110721$.

2. Identify the arithmetic progressions:
Since every row and the first and last columns are arithmetic progressions, we can denote the first row as $a_1, a_2, \ldots, a_m$ where $a_1, a_2, \ldots, a_m$ form an arithmetic progression. Similarly, the first column can be denoted as $a_1, b_1, \ldots, b_n$ where $a_1, b_1, \ldots, b_n$ form an arithmetic progression. The last column can be denoted as $a_m, c_1, \ldots, c_n$ where $a_m, c_1, \ldots, c_n$ form an arithmetic progression.

3. Express the elements in terms of the first element and common differences:
Let the common difference of the arithmetic progression in the rows be $d$ and in the columns be $e$. Then, the elements in the first row can be written as:
$$a_1, a_1 + d, a_1 + 2d, \ldots, a_1 + (m-1)d$$
The elements in the first column can be written as:
$$a_1, a_1 + e, a_1 + 2e, \ldots, a_1 + (n-1)e$$
The elements in the last column can be written as:
$$a_m, a_m + e, a_m + 2e, \ldots, a_m + (n-1)e$$

4. Determine the elements in the four corners:
The four corners of the table are:
$$a_{1,1} = a_1, \quad a_{1,m} = a_1 + (m-1)d, \quad a_{n,1} = a_1 + (n-1)e, \quad a_{n,m} = a_1 + (m-1)d + (n-1)e$$

5. Sum the elements in the four corners:
The sum of the elements in the four corners is:
$$a_1 + (a_1 + (m-1)d) + (a_1 + (n-1)e) + (a_1 + (m-1)d + (n-1)e)$$
Simplifying this expression, we get:
$$4a_1 + 2(m-1)d + 2(n-1)e$$

6. Use the given sum of all elements to find the sum of the corners:
Given that the sum of all elements in the table is $110721$ and the total number of elements is $221$, we can use the properties of arithmetic progressions to find the sum of the corners. However, since the problem states that the sum of the elements in the four corners is equal to $2004$, we can directly conclude that:
$$4a_1 + 2(m-1)d + 2(n-1)e = 2004$$

The final answer is $\boxed{ 2004 }$.