For how many integers $0\leq n < 2013$, is $n^4+2n^3-20n^2+2n-21$ divisible by $2013$?

$ 
\textbf{(A)}\ 6
\qquad\textbf{(B)}\ 8
\qquad\textbf{(C)}\ 12 
\qquad\textbf{(D)}\ 16
\qquad\textbf{(E)}\ \text{None of above}
$
To solve the problem, we need to determine how many integers \(0 \leq n < 2013\) make the polynomial \(n^4 + 2n^3 - 20n^2 + 2n - 21\) divisible by 2013. 

First, we factorize the polynomial:
\[ n^4 + 2n^3 - 20n^2 + 2n - 21 = (n^2 + 1)(n^2 + 2n - 21) \]

Next, we factorize 2013:
\[ 2013 = 3 \times 11 \times 61 \]

We need to find \(n\) such that:
\[ (n^2 + 1)(n^2 + 2n - 21) \equiv 0 \pmod{2013} \]

Since \(2013 = 3 \times 11 \times 61\), we need to check the divisibility by each of these prime factors.

### Step 1: Divisibility by 3
For \(n^2 + 1 \equiv 0 \pmod{3}\):
\[ n^2 \equiv -1 \pmod{3} \]
This is impossible because the quadratic residues modulo 3 are 0 and 1. Therefore, \(n^2 + 1\) is never divisible by 3.

For \(n^2 + 2n - 21 \equiv 0 \pmod{3}\):
\[ n^2 + 2n - 21 \equiv n^2 + 2n \equiv 0 \pmod{3} \]
\[ n(n + 2) \equiv 0 \pmod{3} \]
This gives us two cases:
\[ n \equiv 0 \pmod{3} \]
\[ n \equiv 1 \pmod{3} \]

### Step 2: Divisibility by 11
For \(n^2 + 1 \equiv 0 \pmod{11}\):
\[ n^2 \equiv -1 \pmod{11} \]
The solutions to this are:
\[ n \equiv 5 \pmod{11} \]
\[ n \equiv 6 \pmod{11} \]

For \(n^2 + 2n - 21 \equiv 0 \pmod{11}\):
\[ n^2 + 2n - 21 \equiv n^2 + 2n + 1 \equiv 0 \pmod{11} \]
\[ (n + 1)^2 \equiv 22 \pmod{11} \]
\[ (n + 1)^2 \equiv 0 \pmod{11} \]
\[ n + 1 \equiv 0 \pmod{11} \]
\[ n \equiv 10 \pmod{11} \]

### Step 3: Divisibility by 61
For \(n^2 + 1 \equiv 0 \pmod{61}\):
\[ n^2 \equiv -1 \pmod{61} \]
The solutions to this are:
\[ n \equiv 23 \pmod{61} \]
\[ n \equiv 38 \pmod{61} \]

For \(n^2 + 2n - 21 \equiv 0 \pmod{61}\):
\[ n^2 + 2n - 21 \equiv 0 \pmod{61} \]
Solving the quadratic equation:
\[ n = \frac{-2 \pm \sqrt{4 + 84}}{2} \]
\[ n = \frac{-2 \pm \sqrt{88}}{2} \]
\[ n = \frac{-2 \pm 2\sqrt{22}}{2} \]
\[ n = -1 \pm \sqrt{22} \]
The solutions modulo 61 are:
\[ n \equiv 10 \pmod{61} \]
\[ n \equiv 51 \pmod{61} \]

### Combining the Solutions
We need to find common solutions for \(n\) modulo 3, 11, and 61. Using the Chinese Remainder Theorem, we combine the solutions:

1. \(n \equiv 0 \pmod{3}\)
2. \(n \equiv 1 \pmod{3}\)
3. \(n \equiv 5 \pmod{11}\)
4. \(n \equiv 6 \pmod{11}\)
5. \(n \equiv 10 \pmod{11}\)
6. \(n \equiv 23 \pmod{61}\)
7. \(n \equiv 38 \pmod{61}\)
8. \(n \equiv 10 \pmod{61}\)
9. \(n \equiv 51 \pmod{61}\)

By solving these congruences, we find the distinct solutions for \(n\) in the range \(0 \leq n < 2013\).

The final answer is \(\boxed{6}\).