This is equivalent to the existence of $a,b\in\mathbb Z$ such that $\operatorname{gcd}(a+i,b+j)>1$ for all $i,j\in\{0,1,\dots,n-1\}$. However, we can simply force such such $a$ and $b$ using the Chinese Remainder Theorem. let $p_k$ be the $k$th prime, then we look at the $2n^2$ congruences \[a+i\equiv b+j\equiv 0\pmod{p_{1+j+ni}}\]for all $i,j\in\{0,1,\dots,n-1\}$. By CRT, this has a solution $(a,b)\in\mathbb Z^2$.