All integers are in a row.

The case $n=2$ is best handled separately. The matrix of elements must be of the form $\left(\begin{array}{cc} 1 +2a & 1 - 2a + 4b \\ 1 - 2a + 4c & 1 + 2a + 4d \end{array}\right)$ for integers $a,b,c,d$ , and it is easy to calculate that $(R_1+R_2)-(C_1+C_2) = 16(c-b)d$ is always a multiple of $16$ , but not necessarily a multiple of $32$ .

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Suppose now that $n \ge 3$ . We can write the elements of the matrix as $a_{ij} \; = \; 1 + nb_{ij} \hspace{2cm} 1 \le i,j \le n$ where $b_{ij}$ are integers. We also must have integers $\alpha_i,\beta_i$ for $1 \le i \le n$ such that $\sum_{j=1}^n b_{ij} \; = \; n\alpha_i \hspace{1cm} \sum_{j=1}^n b_{ji} \; = \; n\beta_i \hspace{2cm} 1 \le i \le n$

Let us define the quantities $B_k \; = \; \sum_{ij}b_{ij}^k \hspace{2cm} k \in \mathbb{N}$ as well as $\begin{array}{rclcrcl} X_{1i} & \; =\; & {\displaystyle \sum_{j_1<j_2}b_{ij_1}b_{ij_2}} & & X_{2j} & \; = \; & {\displaystyle \sum_{i_1 < i_2}b_{i_1j}b_{i_2j}} \\ Y_{1i} & \;=\; & {\displaystyle \sum_{j_1<j_2<j_3}b_{ij_1}b_{ij_2}b_{ij_3}} & & Z_{2j} & \; =\; & {\displaystyle \sum_{i_1<i_2<i_3}b_{i_1j}b_{i_2j}b_{i_3j}} \end{array}$

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First note that $n\sum_i\alpha_i \; = \; B_1 \; = \; n\sum_j\beta_j$ so that $Q \; = \; \sum_j(\alpha_j^2 - \beta_j^2) \; = \; 2\sum_{i<j}(\beta_i\beta_j - \alpha_i\alpha_j)$ is even. But then $2X_{1i} \; = \; \; = \; \sum_{j_1 \neq j_2}b_{ij_1}b_{ij_2} \; = \; \left(\sum_jb_{ij}\right)^2 - \sum_jb_{ij}^2 \; = \; n^2\alpha_i^2 - \sum_jb_{ij}^2$ so that $2\sum_iX_{1i} \; = \; n^2 \sum_i\alpha_i^2 - B_2$ Similarly $2\sum_jX_{2j} \; = \; n^2\sum_j\beta_j^2 - B_2$ and hence $2\sum_iX_{1i} - 2\sum_jX_{2j} \; = \; Qn^2$ so that $\sum_iX_{1i} \; \equiv \; \sum_jX_{2j} \hspace{2cm} (\mathrm{mod}\; n^2)$

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Next note that $\begin{aligned} \left(\sum_jb_{ij}\right)^3 & = \; \sum_jb_{ij}^3 + 3\sum_{j_1 \neq j_2}b_{ij_1}^2b_{ij_2} + 6Y_{1i} \\ & = \; -2\sum_j b_{ij}^3 + 3\left(\sum_jb_{ij}^2\right)\left(\sum_jb_{ij}\right) + 6Y_{1i} \\ & = \; -2\sum_j b_{ij}^3 + 3\left(\sum_j b_{ij}\right)^3 - 6\left(\sum_{j_1<j_2}b_{ij_1}b_{ij_2}\right)\left(\sum_jb_{ij}\right) + 6Y_{1i} \\ 3Y_{1i} & = \; \sum_ib_{ij}^3 - n^3\alpha_i^3 + 3n\alpha_iX_{1i} \end{aligned}$ so that $3\sum_iY_{1i} \; = \; B_3 - n^3\sum_i\alpha_i^3 + 3n\sum_i\alpha_iX_{1i}$ Similarly $3\sum_jY_{2j} \; = \; B_3 - n^3\sum_j\beta_j^3 + 3n\sum_j\beta_jX_{2j}$ and hence $3\left(\sum_iY_{1i} - \sum_jY_{2j}\right) \; = \; n^3\sum_j(\beta_j^3 - \alpha_j^3) + 3n\sum_j(\alpha_jX_{1j} - \beta_jX_{2j})$ from which it is clear that $\sum_iY_{1i} \; \equiv \; \sum_jY_{2j} \hspace{2cm} (\mathrm{mod}\; n)$ even when $n$ is divisible by $3$ .

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We now note that $R_i \; = \; \prod_ja_{ij} \; = \; \prod_j(1 + nb_{ij}) \equiv \; 1 + n\sum_jb_{ij} + n^2X_{1j} +n^3Y_{1i} \hspace{2cm}(\mathrm{mod}\; n^4)$ so that $\sum_i R_i \; = \; n + nB_1 + n^2\sum_iX_{1i} + n^3\sum_iY_{1i} \hspace{2cm} (\mathrm{mod}\;n^4)$ Similarly $\sum_j C_j \; = \; n + nB_1 + n^2\sum_jX_{2j} + n^3\sum_jY_{2j} \hspace{2cm} (\mathrm{mod}\;n^4)$ and hence we deduce that $\sum_i R_i \; \equiv \; \sum_j C_j \hspace{2cm} (\mathrm{mod}\;n^4)$

Thus $\sum_iR_i$ and $\sum_jC_j$ are congruent modulo $n^4$ for any integer $n \ge 2$ . Since the two quantities are not always congruent modulo $n^5$ , as can be seen by the $n=2$ case, we see that the correct answer is $\boxed{n^4}$ .