 Solution

Choose distinct primes $p_1,\dots ,p_{1000}$. The moduli $p_k^2$ are pairwise coprime, so the Chinese remainder theorem supplies an integer $N$ satisfyingConsequentlyso $p_k\mid N+k$ but $p_k^2\nmid N+k$. By the definition of a squarefull number, $N+k$ is therefore not squarefull. Adding a sufficiently large multiple of ${\prod }_k{p}_k^2$ makes $N$ positive without changing any congruence. The integers $N+1,\dots ,N+1000$ are the required consecutive integers.