Because there are infinite many prime numbers, it is always possible to select $n$ distinct prime numbers $p_1$, $p_2$, $\cdots$, $p_n$. Consider the following system of congruence $$\left\{ \begin{array}{rclclrl} x + 1 &\equiv 0 &\pmod{p_1^2} &\implies & x\equiv & -1 &\pmod{p_1^2} \\ x +2 & \equiv 0 & \pmod{p_2^2} &\implies & x\equiv & -2 &\pmod{p_2^2} \\ & \cdots \\ x+n & \equiv  0 &\pmod{p_n^2} &\implies & x\equiv & -n&\pmod{p_n^2}\end{array} \right. $$

Because all the modulo are pairwise co-prime, by the Chinese Remainder Theorem, this system is solvable. Therefore, there exist $n$ consecutive numbers $(x + i)$, $(1 \le i \le n)$, where $p_i^2\mid x + i$.