IMO Shortlist

Let $A_{i, j}$ be the entry in the $i^{\text{th}}$ row and the $j^{\text{th}}$ column; let $P$ be the product of all $n^{2}$ entries. For convenience, denote $a_{i, j} = A_{i, j} - 1$ and $r_{i} = R_{i} - 1$. We show that
$$\sum_{i=1}^{n} R_{i} \equiv (n-1) + P \quad (\bmod\ n^{4}) \tag{1}$$
Due to symmetry of the problem conditions, the sum of all the $C_{j}$ is also congruent to $(n-1) + P$ modulo $n^{4}$, whence the conclusion.
By condition $(i)$, the number $n$ divides $a_{i, j}$ for all $i$ and $j$. So, every product of at least two of the $a_{i, j}$ is divisible by $n^{2}$, hence
$R_{i} = \prod_{j=1}^{n} (1 + a_{i, j}) = 1 + \sum_{j=1}^{n} a_{i, j} + \sum_{1 \leqslant j_{1} < j_{2} \leqslant n} a_{i, j_{1}} a_{i, j_{2}} + \cdots \equiv 1 + \sum_{j=1}^{n} a_{i, j} \equiv 1 - n + \sum_{j=1}^{n} A_{i, j} \ (\bmod\ n^{2})$
for every index $i$. Using condition (ii), we obtain $R_{i} \equiv 1\ (\bmod\ n^{2})$, and so $n^{2} \mid r_{i}$.
Therefore, every product of at least two of the $r_{i}$ is divisible by $n^{4}$. Repeating the same argument, we obtain
$$P = \prod_{i=1}^{n} R_{i} = \prod_{i=1}^{n} (1 + r_{i}) \equiv 1 + \sum_{i=1}^{n} r_{i} \quad (\bmod\ n^{4}),$$
whence
$$\sum_{i=1}^{n} R_{i} = n + \sum_{i=1}^{n} r_{i} \equiv n + (P - 1) \quad (\bmod\ n^{4})$$
as desired.

We present a more straightforward (though lengthier) way to establish (1). We also use the notation of $a_{i, j}$.
By condition (i), all the $a_{i, j}$ are divisible by $n$. Therefore, we have
$$\begin{aligned} P = \prod_{i=1}^{n} \prod_{j=1}^{n} (1 + a_{i, j}) &\equiv 1 + \sum_{(i, j)} a_{i, j} + \sum_{(i_{1}, j_{1}), (i_{2}, j_{2})} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} \\ &+ \sum_{(i_{1}, j_{1}), (i_{2}, j_{2}), (i_{3}, j_{3})} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}} \quad (\bmod\ n^{4}) \end{aligned}$$
where the last two sums are taken over all unordered pairs/triples of pairwise different pairs $(i, j)$; such conventions are applied throughout the solution.
Similarly,
$$\sum_{i=1}^{n} R_{i} = \sum_{i=1}^{n} \prod_{j=1}^{n} (1 + a_{i, j}) \equiv n + \sum_{i} \sum_{j} a_{i, j} + \sum_{i} \sum_{j_{1}, j_{2}} a_{i, j_{1}} a_{i, j_{2}} + \sum_{i} \sum_{j_{1}, j_{2}, j_{3}} a_{i, j_{1}} a_{i, j_{2}} a_{i, j_{3}} \quad (\bmod\ n^{4}) .$$
Therefore,
$$\begin{aligned} P + (n-1) - \sum_{i} R_{i} &\equiv \sum_{\substack{(i_{1}, j_{1}), (i_{2}, j_{2}) \\ i_{1} \neq i_{2}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} \\ &+ \sum_{\substack{(i_{1}, j_{1}), (i_{2}, j_{2}), (i_{3}, j_{3}) \\ i_{1} \neq i_{2} \neq i_{3} \neq i_{1}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}} \\ &+ \sum_{\substack{(i_{1}, j_{1}), (i_{2}, j_{2}), (i_{3}, j_{3}) \\ i_{1} \neq i_{2} = i_{3}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}} \quad (\bmod\ n^{4}) . \end{aligned}$$
We show that in fact each of the three sums appearing in the right-hand part of this congruence is divisible by $n^{4}$; this yields (1). Denote those three sums by $\Sigma_{1}, \Sigma_{2}$, and $\Sigma_{3}$ in order of appearance. Recall that by condition (ii) we have
$$\sum_{j} a_{i, j} \equiv 0 \quad (\bmod\ n^{2}) \quad \text{for all indices } i$$
For every two indices $i_{1} < i_{2}$ we have
$$\sum_{j_{1}} \sum_{j_{2}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} = \left(\sum_{j_{1}} a_{i_{1}, j_{1}}\right) \cdot \left(\sum_{j_{2}} a_{i_{2}, j_{2}}\right) \equiv 0 \quad (\bmod\ n^{4})$$
since each of the two factors is divisible by $n^{2}$. Summing over all pairs $(i_{1}, i_{2})$ we obtain $n^{4} \mid \Sigma_{1}$.
Similarly, for every three indices $i_{1} < i_{2} < i_{3}$ we have
$$\sum_{j_{1}} \sum_{j_{2}} \sum_{j_{3}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}} = \left(\sum_{j_{1}} a_{i_{1}, j_{1}}\right) \cdot \left(\sum_{j_{2}} a_{i_{2}, j_{2}}\right) \cdot \left(\sum_{j_{3}} a_{i_{3}, j_{3}}\right)$$
which is divisible even by $n^{6}$. Hence $n^{4} \mid \Sigma_{2}$.
Finally, for every indices $i_{1} \neq i_{2} = i_{3}$ and $j_{2} < j_{3}$ we have
$$a_{i_{2}, j_{2}} \cdot a_{i_{2}, j_{3}} \cdot \sum_{j_{1}} a_{i_{1}, j_{1}} \equiv 0 \quad (\bmod\ n^{4})$$
since the three factors are divisible by $n, n$, and $n^{2}$, respectively. Summing over all 4-tuples of indices $(i_{1}, i_{2}, j_{2}, j_{3})$ we get $n^{4} \mid \Sigma_{3}$.