Step 1: Recall the trailing-zero rule.
The number of zeros at the end of $n!$ equals the count of factors of $5$, namely $Z(n)=\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{25}\right\rfloor+\left\lfloor\frac{n}{125}\right\rfloor+\left\lfloor\frac{n}{625}\right\rfloor+\cdots$
Step 2: Estimate the size of $n$.
Since roughly $Z(n)\approx\frac{n}{4}$, for $1000$ zeros $n$ is near $4000$.
Step 3: Test $n=4000$.
$\left\lfloor\frac{4000}{5}\right\rfloor+\left\lfloor\frac{4000}{25}\right\rfloor+\left\lfloor\frac{4000}{125}\right\rfloor+\left\lfloor\frac{4000}{625}\right\rfloor+\left\lfloor\frac{4000}{3125}\right\rfloor=800+160+32+6+1=999$.
Step 4: Move up slightly.
One more multiple of $5$ is needed, so check values just above $4000$.
Step 5: Test $n=4005$.
$801+160+32+6+1=1000$. So $n=4005$ gives $1000$ zeros, and this count stays $1000$ up to $4009$, since no new factor of $5$ enters until $4010$.
Step 6: Choose the listed value.
Among the options, $4009$ lies in this range and gives exactly $1000$ zeros, which is option (3).
\[ \boxed{4009} \]