17. If $P(x)=\left(x-1^{2}\right)\left(x-2^{2}\right) \cdots\left(x-100^{2}\right)$, then there are $(\quad)$ integers $n$, such that $P(n) \leqslant 0$.
(A) 4900
(B) 4950
(C) 5000
(D) 5050
(E) 5100
17. E.
$$
\begin{array}{c}
\text { Given } P(n)=\left(n-1^{2}\right)\left(n-2^{2}\right) \cdots\left(n-100^{2}\right) \leqslant 0 \\
\Rightarrow n \in\left[1^{2}, 2^{2}\right] \cup\left[3^{2}, 4^{2}\right] \cup \cdots \cup \\
{\left[99^{2}, 100^{2}\right] .}
\end{array}
$$

Notice that, in the set
$$
\left[1^{2}, 2^{2}\right] \cup\left[3^{2}, 4^{2}\right] \cup \cdots \cup\left[99^{2}, 100^{2}\right]
$$

the number of integers is
$$
\begin{aligned}
N= & 2^{2}-1^{2}+1+4^{2}-3^{2}+1+\cdots+ \\
& 100^{2}-99^{2}+1 \\
= & (2-1)(2+1)+1+(4-3)(4+3)+ \\
& 1+\cdots+(100-99)(100+99)+1 \\
= & 1+2+3+4+\cdots+99+100+50 \\
= & 5100 .
\end{aligned}
$$