1. Statement I: $\(n^2\)$ is divisible by 600

First, we find the prime factorization of 600 :

$$
\(600=6 \times 100=(2 \times 3) \times\left(2^2 \times 5^2\right)=2^3 \times 3^1 \times 5^2\)
$$


Now, let's look at $\(n^2\)$ :

$$
\(\begin{gathered}
n=2^1 \times 3^1 \times 5^1 \times 7^1 \times 11^1 \times 13^1 \times 17^1 \\
n^2=2^2 \times 3^2 \times 5^2 \times 7^2 \times 11^2 \times 13^2 \times 17^2
\end{gathered}\)
$$


For $\(n^2\)$ to be divisible by 600 , it must contain at least as many of each prime factor as 600 does:
- $\(n^2\)$ has $\(2^2\)$, but 600 requires $\(2^3\)$.

Since $\(2^2<2^3, n^2\)$ is not divisible by 600 . (Statement I is False)
2. Statement II: $\(n+19\)$ is divisible by $\(\mathbf{1 9}\)$

For $\(n+19\)$ to be divisible by 19 , $n$ must also be divisible by 19 (because $\(n+19=19\left(\frac{n}{19}+1\right)\)$ ).
$n$ is the product of all prime numbers from 2 to 17.19 is a prime number that is not in this list.
Therefore, $n$ is not divisible by 19 .
Consequently, $\(n+19\)$ is not divisible by 19 . (Statement II is False)

3. Statement III: $\(\frac{n+4}{2}\)$ is even

We can simplify the expression:

$$
\(\frac{n+4}{2}=\frac{n}{2}+\frac{4}{2}=\frac{n}{2}+2\)
$$


Substitute the factors of $n$ :

$$
\(\frac{n}{2}=\frac{2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17}{2}=3 \times 5 \times 7 \times 11 \times 13 \times 17\)
$$


The resulting number is a product of odd numbers, which is always odd.

$$
\(\text { Odd }+\operatorname{Even}(2)=\text { Odd }\)
$$


Therefore, $\(\frac{n+4}{2}\)$ is odd, not even. (Statement III is False)

Conclusion
Since none of the statements are true, the correct answer is $\(\mathbf{E}\)$. None of the above.