27. Let $i, j \in \mathbf{Z}, -n \leqslant i, j \leqslant n$, then $p_{ij}$ represents $(2n+1)^2$ different primes. By the Chinese Remainder Theorem, there exist $a, b$ that satisfy the congruence systems
$a \equiv i \pmod{p_{ij}}, -n \leqslant i, j \leqslant n, b \equiv j \pmod{p_{ij}}, -n \leqslant i, i \leqslant n$, then $(a, b)$ is an integer point that satisfies the conditions. Because if the distance between $(x, y)$ and $(a, b)$ is $\leqslant n$, then
$$\sqrt{(a-x)^2 + (b-y)^2} \leqslant n$$

which implies
$$(a-x)^2 + (b-y)^2 \leqslant n^2$$. This leads to
$$|a-x| \leqslant n, |b-y| \leqslant n$$

Assume without loss of generality that $a-x=i, b-y=j$, i.e., $x=a-i, y=b-j, -n \leqslant i, j \leqslant n$. By the choice of $a, b$, it follows that $x, y$ are multiples of $p_{ij}$, hence $(x, y)$ is not a primitive integer point.