$A$ is an $l\times m$ matrix with integer entries and each column of which contains at least one negative entry. $y$ is a column matrix with integer entries of length $l$. Define the set of sequence $X = \{(x_1,x_2,\cdots,x_n)\text{ for some natural number }n\}$ as follows. $x_i$ is a column matrix of length $m$ with all zero but one natural number entry $\forall 1\le i\le n$ for some natural number $n$. Note different $x$'s have different $n$'s. For any $x\in X$, let $s(x):= (s_1,s_2,\cdots,s_n)$ where $s_1 := y+Ax_1$ and $s_{i+1} := s_i+Ax_{i+1}$, and the entries of $s_i$ have to be all non-negative, $\forall i\in [1,2,\cdots, n-1]$. Let $\alpha(x)$ be the cardinality of $x\in X$. In other words, $\alpha(x)$ is the index of the last element of the sequence $x$.
Given a positive integer $j$, find $x_{\text{max}}$ that maximizes the $j$'th entry of $s_{\alpha(x)}$ of $s(x)$ for all $x\in X$.
The original formulation of the problem is not well posed as pointed out by Hans Engler in his answer. I have now modified the formulation of the problem.
Would we solve this problem using say, integer programming, linear algebra over finite field, or group theory?
Note that $s_j = A(x_1 + x_2 + \dots + x_j)$.You can make the entries of such a vector as large as you want by choosing a sufficiently large $x_1 = x_2 = x_3 = \dots$ or by making $n$ sufficiently large.
Your problem is not well posed.
More generally: Since $s_j = A (\sum_{i \le j} x_i)$, once you have a sequence $x \in X$ with a positive entry somewhere in $s_{\alpha(x)}$, you can just create a new sequence $y \in X$ by repeating all elements of $x$ as many times as you want, say $M$ times. Then the new $s_{\alpha(y)}$ equals $M s_{\alpha(x)}$.
Your objective function is not bounded above.