The problem essentially asks for a magic square that satisfies an additional constant-sum property along the wrap-around diagonals.
Suppose that such an arrangement of $1, 2, \ldots, 100$ is possible. Since the sum of all entries is $\frac{1}{2} \cdot 100 \cdot 101$, we get that $S = 505$ is an odd number. We partition the cells $C(i,j)$ into four sets: Set $A$ contains the cells with $i$ and $j$ both odd; set $B$ contains the cells with odd $i$ and even $j$; set $C$ contains the cells with even $i$ and odd $j$; and set $D$ contains the remaining cells with $i$ and $j$ both even. We denote the sum of all entries in $A, B, C, D$ by $S_A, S_B, S_C, S_D$, respectively.
* Since $A$ and $B$ contain all cells in the odd rows, we get $S_A + S_B = 5S$.
* Since $B$ and $D$ contain all cells in the even columns, we get $S_B + S_D = 5S$.
* Since $A$ and $D$ contain all cells $C(i,j)$ with even $i-j$, we get $S_A + S_D = 5S$.
Adding up these three equations yields $2(S_A + S_B + S_D) = 15S$. Since in this equation the left hand side is even and the right hand side is odd, we have the desired contradiction.