Suppose that such an arrangement of $1,2,\ldots,100$ is possible. Since the sum of all entries is $\frac12\cdot100\cdot101$, 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 $B$ together contain all cells in the odd rows, we get $S_A+S_B=5\,S$. Since $B$ and $D$ together contain all cells in the even columns, we get $S_B+S_D=5\,S$. Since $A$ and $D$ together contain all cells $C(i,j)$ with even $i-j$, we get $S_A+S_D=5\,S$. Adding up these three equations yields $2(S_A+S_B+S_D)=15\,S$. Since in this equation the left hand side is even and the right hand side is odd, we have the desired contradiction.