First let us prove the following lemma.

Lemma. There are $l$ consecutive positive integers, among which there are no numbers with the degree less than $t$ ($t \ge 2$).

Proof is conducted by Mathematical Induction with respect to $t$.

Base for $t=2$. The statement is equivalent to the fact that there are $l$ consecutive positive integers, none of which is prime. So it is sufficient to take $(l+1)!+2, (l+1)!+3, \ldots, (l+1)!(l+1)$.

Assume that we have proved the statement for $2, 3, \ldots, (t-1)$. Let us prove it for $t$. Suppose for $(t-1)$ $x+1, \ldots, x+l$ are the numbers with the degree not less than $(t-1)$. Now consider the numbers $(x+l)!+x+1, \ldots, (x+l)!+x+l$. Obviously, each of them has the degree not less than $t$ and there are $l$ of them. Thus, the lemma is proved.

Note that $2^{11} = 2048 > 2016$. So among the first $2016$ positive integers all numbers have the degree less than $11$. Let $w(x)$ be the amount of numbers with the degree less than $11$ among the numbers $x+1, \ldots, x+2016$. Hence, $w(0) = 2016$. At the same time we have shown that there are $2016$ consecutive positive integers which do not include any of the numbers with the degree less than $11$. So there exists a positive integer $y$ such that $w(y)=0$. It is also clear that $w(z)-1 \le w(z+1) \le w(z)+1$. Thus, as $w(z)$ takes only positive integer values or zero, in $2$ consecutive points differs by not less than $1$ and takes the values $0$ and $2016$, it has to also take all the intermediate values, therefore $1000$ as well.