Let $A$ be $(n-1) \times n$ matrix that is of the following form: $$\left( \begin{array}{ccc} n-1 & 1 & 0 &.... & ....\\ 0 & n-2 & 2 & .... & ....\\ 0 & 0 & n-3 & 3 & .... \\ \vdots & \ddots &\ddots \end{array} \right)$$.
All numbers in $A$ are non-negative integers.
In $(n-1) \times 1$ vector $b$, all numbers do not exceed $n^{(n^4)}$.
Is there any $n$ such that there exists unique vector solution $x$ for $Ax = b$ where $x$ only consists of non-negative integers?
The null space of $A$ is spanned by vector $v$ with $v_i = (-1)^{i-1} {n-1 \choose i} v_1$ for $i \le n-1$, $v_n = 0$. The difference between any two integer solutions of $Ax=b$ is an integer multiple of $v$. Take, e.g. $X$ with only two nonzero entries, $X_i = 1$ and $X_j = 1$, where $i$ is even and $j$ is odd, $1 < i,j < n-1$, and $b = A X$. Then $X$ is the unique nonnegative integer solution of $AX = b$.