1. We start with the expression $\pm 1 \pm 2 \pm 3 \pm \cdots \pm 2012$. We need to determine the number of possible remainders when this expression is divided by 2012.

2. Consider the sum of the sequence $1 + 2 + 3 + \cdots + 2012$. The sum of the first $n$ natural numbers is given by:
$$S = \frac{n(n+1)}{2}$$
For $n = 2012$, we have:
$$S = \frac{2012 \times 2013}{2} = 2012 \times 1006.5$$

3. Since we are placing either a $+$ or $-$ sign in front of each number, the sum can range from $-S$ to $S$. However, we are interested in the remainders when this sum is divided by 2012.

4. Notice that each number $k$ in the sequence can be paired with $k-1$ to form either $k - (k-1)$ or $(k-1) - k$, which simplifies to $1$ or $-1$. This means we can form pairs that contribute $\pm 1$ to the total sum.

5. There are 1006 such pairs (since $2012/2 = 1006$). Therefore, the possible sums are all even numbers from $-2012 \times 1006$ to $2012 \times 1006$.

6. To find the possible remainders when divided by 2012, we need to consider the sums modulo 2012. Since the sums are even, the remainders must also be even.

7. The possible even remainders modulo 2012 are $0, 2, 4, \ldots, 2010$. This forms an arithmetic sequence with the first term $a = 0$ and common difference $d = 2$.

8. The number of terms in this sequence is given by:
$$n = \frac{2010 - 0}{2} + 1 = 1006$$

9. Therefore, there are 1006 possible even remainders when the expression is divided by 2012.

The final answer is $\boxed{1006}$