For how many integers $n$ with $1 \le n \le 2012$ is the product
\[
  \prod_{k=0}^{n-1} \left( \left( 1 + e^{2 \pi i k / n} \right)^n + 1 \right)
\]
equal to zero?
1. We start by analyzing the given product:
   \[
   \prod_{k=0}^{n-1} \left( \left( 1 + e^{2 \pi i k / n} \right)^n + 1 \right)
   \]
   We need to determine for which integers \( n \) this product equals zero.

2. Consider the term inside the product:
   \[
   \left( 1 + e^{2 \pi i k / n} \right)^n + 1
   \]
   For this term to be zero, we need:
   \[
   \left( 1 + e^{2 \pi i k / n} \right)^n = -1
   \]

3. Let \( \theta = \frac{2 \pi k}{n} \). Then:
   \[
   1 + e^{i \theta} = 1 + \cos \theta + i \sin \theta
   \]
   We need:
   \[
   \left( 1 + \cos \theta + i \sin \theta \right)^n = -1
   \]

4. The magnitude of \( 1 + \cos \theta + i \sin \theta \) must be 1:
   \[
   |1 + \cos \theta + i \sin \theta| = 1
   \]
   This implies:
   \[
   (1 + \cos \theta)^2 + \sin^2 \theta = 1
   \]
   Simplifying, we get:
   \[
   1 + 2 \cos \theta + \cos^2 \theta + \sin^2 \theta = 1
   \]
   \[
   1 + 2 \cos \theta + 1 = 1
   \]
   \[
   2 \cos \theta = -1
   \]
   \[
   \cos \theta = -\frac{1}{2}
   \]

5. The angles \( \theta \) that satisfy \( \cos \theta = -\frac{1}{2} \) are:
   \[
   \theta = \frac{2\pi}{3} \quad \text{or} \quad \theta = \frac{4\pi}{3}
   \]
   Therefore, \( e^{i \theta} \) must be either \( e^{2\pi i / 3} \) or \( e^{4\pi i / 3} \).

6. For \( e^{2\pi i k / n} \) to be \( e^{2\pi i / 3} \) or \( e^{4\pi i / 3} \), \( n \) must be a multiple of 3. Hence, \( n = 3m \) for some integer \( m \).

7. Additionally, if \( n \) is even, then:
   \[
   \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right)^n = e^{n \pi i / 3} \quad \text{or} \quad \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right)^n = e^{-n \pi i / 3}
   \]
   Both of these expressions are 1, not -1, so \( n \) must be odd.

8. Therefore, \( n \) must be of the form \( 3m \) where \( m \) is an odd integer.

9. We need to count the number of such \( n \) in the range \( 1 \leq n \leq 2012 \). Let \( n = 3m \). Then:
   \[
   1 \leq 3m \leq 2012 \implies 1 \leq m \leq \frac{2012}{3} \approx 670.67
   \]
   So, \( m \) ranges from 1 to 670.

10. Since \( m \) must be odd, the number of odd integers from 1 to 670 is:
    \[
    \left\lfloor \frac{670 + 1}{2} \right\rfloor = 335
    \]

The final answer is \(\boxed{335}\)