Problem 1224

1. Sum of all entries:
The sum of all integers from 1 to 100 is given by the formula for the sum of an arithmetic series:
$$\sum_{k=1}^{100} k = \frac{100 \cdot 101}{2} = 5050$$
Since there are 10 rows and 10 columns, the sum of the entries in each row and each column must be the same, denoted by $S$. Therefore, we have:
$$10S = 5050 \implies S = 505$$

2. Partitioning the cells:
We partition the cells $C(i, j)$ into four sets:
- Set $A$ contains cells where both $i$ and $j$ are odd.
- Set $B$ contains cells where $i$ is odd and $j$ is even.
- Set $C$ contains cells where $i$ is even and $j$ is odd.
- Set $D$ contains cells where both $i$ and $j$ are even.

3. Sum of entries in each set:
Let $S_A, S_B, S_C, S_D$ denote the sum of all entries in sets $A, B, C, D$ respectively.

4. Equations based on the conditions:
- Since $A$ and $B$ together contain all cells in the odd rows, we have:
$$S_A + S_B = 5S$$
- Since $B$ and $D$ together contain all cells in the even columns, we have:
$$S_B + S_D = 5S$$
- Since $A$ and $D$ together contain all cells $C(i, j)$ with even $i - j$, we have:
$$S_A + S_D = 5S$$

5. Adding the equations:
Adding the three equations, we get:
$$(S_A + S_B) + (S_B + S_D) + (S_A + S_D) = 15S$$
Simplifying, we obtain:
$$2(S_A + S_B + S_D) = 15S$$
Since $S = 505$, which is an odd number, the right-hand side of the equation is odd:
$$15S = 15 \cdot 505 = 7575$$
The left-hand side, $2(S_A + S_B + S_D)$, is even because it is twice a sum of integers. This leads to a contradiction because an even number cannot equal an odd number.

Therefore, the assumption that such an arrangement is possible leads to a contradiction.

$\blacksquare$