= Solution
Choose distinct <primes> $p_1,\ldots,p_{1000}$. The moduli $p_k^2$ are pairwise <coprime>, so the <Chinese remainder theorem> supplies an integer $N$ satisfying
$$
N\equiv p_k-k\pmod {p_k^2}
\qquad(1\leq k\leq1000).
$$
Consequently
$$
N+k\equiv p_k\pmod {p_k^2},
$$
so $p_k\mid N+k$ 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 $\prod_kp_k^2$ makes $N$ positive without changing any congruence. \b[The integers $N+1,\ldots,N+1000$ are the required consecutive integers.]
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