Let $A_{i,j}$ be the entry on row $i$ and column $j$. Let $P$ be the product of all $n^2$ entries. Denote $a_{i,j} = A_{i,j} - 1$ and $r_i = R_i - 1$.
By condition (i), the number $n$ divides $a_{i,j}$. So every product of two or more $a_{i,j}$ is divisible by $n^2$, hence
$$R_i = \prod_{j=1}^n (1 + a_{i,j}) \equiv 1 + \sum_{j=1}^n a_{i,j} \equiv 1 - n + \sum_{j=1}^n A_{i,j} \pmod{n^2}$$
for every $i$.
By condition (ii), we have $R_i \equiv 1 \pmod{n^2}$, and so $n^2|r_i$. Therefore, every product of at least two of the $r_i$ is divisible by $n^4$. Thus
$$P = \prod_{i=1}^n (1 + r_i) \equiv 1 + \sum_{i=1}^n r_i \pmod{n^4}$$
whence
$$\sum_{i=1}^n R_i = n + \sum_{i=1}^n r_i \equiv n - 1 + P \pmod{n^4}$$

Due to symmetry of the problem conditions, we also have
$$\sum_{j=1}^n C_{j} \equiv n-1+P \pmod{n^{4}},$$
thus $\sum_{i=1}^n R_i - \sum_{j=1}^n C_j$ is divisible by $n^4$.