Let $V$ be the $\mathbb Q$-linear space with basis $X$, and $V_0$ be the span of the vectors $\sum_kr_{ik}-\sum_jb_{ij}$ for $i< n$, and $\sum_lr_{jl}-\sum_ib_{ij}$$\sum_lc_{jl}-\sum_ib_{ij}$ for $j< n$.
Proof: A grid $G$ induces a linear mapping $\phi_G$ from $V$ into the space $W$ with basisan $e_1,\dots,e_{n^2}$$n^2$-dimensional such that for any $x'\in X$, the $i$th coordinate of $\phi_G(x')$ gives the number of occurrences of the number $i$ in $x'$. We have $\phi_G(V_0)=0$, and $G$ satisfies $x'$ iff $\phi_G(x')$ is the constant vector $\vec 1$. If $x=\sum_i\alpha_ix_i+y$, where $x_i\in S$ and $y\in V_0$, then $\phi_G(x)=\vec\alpha$ for $\alpha:=\sum_i\alpha_i$. The same holds for every grid $G'$ satisfying $S$; in particular, it holds for any valid grid, which has $\phi_{G'}(x)=\vec1$, hence $\alpha=1$. QED
It remains to show that such a valid grid $G$ exists. We can now forget about $S$, and then it is easy to see that every alternating path can be completed to a cycle, hence we may assume $e$ is a cycle. By applying Sudoku permutations and relabelling the sequence, we may assume $p=0$, $i_m=\lfloor m/2\rfloor$, $j_m=\lceil m/2\rceil$ except that $i_q=j_q=j_{g-1}=0$$i_q=j_q=j_{q-1}=0$. We are thus looking for a solution of the following grid:
$$\begin{array}{|ccc|ccc|ccc|ccc|ccc|}
\hline
1&&&2&&&&&&&&&&&&\\\\
\strut&&&&&&&&&&&&&&&\\\\
\strut&&&&&&&&&&&&&&&\\\\
\hline
&&&1&&&2&&&&&&&&&\\\\
&&&&&&&&&&&&&&\cdots&\\\\
&&&&&&&&\ddots&&&&&&&\\\\
\hline
2&&&&&&&&&1&&&&&&\\\\
\strut&&&&&&&&&&&&&&&\\\\
\strut&&&&&&&&&&&&&&&\\\\
\hline
\strut&&&&&&&&&&&&&&&\\\\
\strut&&&&\vdots&&&&&&&&&&&\\\\
\strut&&&&&&&&&&&&&&&\\\\
\hline
\end{array}$$
where the upper part is a $q'\times q'$ subgrid, $q'=q/2$.
If $q'=n$, we can define the gridsolution easily by putting $s_{ijkl}=(k+l,j-i+l)$, where we relabel the numbers $1,\dots,n^2$ by elements of $(\mathbb Z/n\mathbb Z)\times(\mathbb Z/n\mathbb Z)$, identifying $1$ with $(0,0)$ and $2$ with $(0,1)$. In the general case, we define $s_{ijkl}=(k+l+a_{ij}-b_{ij},l+a_{ij})$. It is easy to check that this is a valid Sudoku if the columns of the matrix $A=(a_{ij})$ and the rows of $B=(b_{ij})$ are permutations of $\mathbb Z/n\mathbb Z$. We obtain the wanted pattern if we let $a_{ij}=b_{ij}=j-i\bmod{q'}$ for $i,j< q'$, and extend this in an arbitrary way so that the columns of $A$ and the rows of $B$ are permutations.
We can also describe the minimal dependent sets. Put
$$D_{R,C}=\{r_{ik}:i\in R,k< n\}\cup\{c_{jl}:j\in C,l< n\}\cup\{b_{ij}:(i\in R\land j\notin C)\lor(i\notin R\land j\in C)\}$$
for $R,C\subseteq\{0,\dots,n-1\}$. If $R$ or $C$ is nonempty, so is $D_{R,C}$, and
$$\sum_{i\in R}\Bigl(\sum_kr_{ik}-\sum_jb_{ij}\Bigr)-\sum_{j\in C}\Bigl(\sum_lc_{jl}-\sum_ib_{ij}\Bigr)\in V_0$$
shows that it$D_{R,C}$ is dependent. On the other hand, if $D$ is a dependent set, there is a linear combination
$$\sum_{i\in R}\alpha_i\Bigl(\sum_kr_{ik}-\sum_jb_{ij}\Bigr)-\sum_{j\in C}\beta_j\Bigl(\sum_lc_{jl}-\sum_ib_{ij}\Bigr)\ne0$$$$\sum_i\alpha_i\Bigl(\sum_kr_{ik}-\sum_jb_{ij}\Bigr)-\sum_j\beta_j\Bigl(\sum_lc_{jl}-\sum_ib_{ij}\Bigr)\ne0$$
where all basic vectors with nonzero coefficients come from $D$. If (WLOG) $\alpha:=\alpha_{i_0}\ne0$, put $R=\{i:\alpha_i=\alpha\}$ and $C=\{j:\beta_j=\alpha\}$. Then $R\ne\varnothing$, and $D_{R,C}\subseteq D$.
This also provides an axiomatization of $\models$ by rules of the form $D\smallsetminus\{x\}\models x$, where $x\in D=D_{R,C}$ is minimal. It is easy to see that if $R=\{i\}$ and $C\ne\varnothing$, the rules for $D_{R,C}$ can be derived from the rules for $D_{R,\varnothing}$ and $D_{\varnothing,\{j\}}$ for $j\in C$, hence we can omit these. (Note that the remaining sets $D_{R,C}$ are completeclosed, hence the corresponding rules have to be included in every axiomatization of $\models$.)
