Label the numbers $a_1, a_2, \dots, a_{2012}$. Let $S(i)$ be the sum of the ten consecutive numbers beginning with $a_i$. $S(i)$ is even for all $i$ and $|S(i) - S(i + 1)| \in \{0, 2\}$. If there existed $i, j$ with $S(i)$ and $S(j)$ of opposite sign, the intermediate value theorem would mean $S(k) = 0$ for some $k$. Then the $S(i)$ are either all positive or all negative. We will assume they are all positive, and introduce a minus sign before giving our final answer. If all the $S(i)$ are positive, the sum over all $i$ of $S(i)$ would have to be at least $2(2012) = 4024$. This counts each number $10$ times, so the sum of all the numbers must be at least $\lceil 4024/10 \rceil = 403$. Of course, the sum is even, so the minimum possible total sum is $404$. This can be attained by setting $a_i = 1$ if $i \equiv 1, 2, 3, 4, 5, 6 \pmod{10}, i \le 2010$, $a_i = -1$ if $i \equiv 7, 8, 9, 0 \pmod{10}, i \le 2010$, and $a_{2011} = a_{2012} = 1$. Every even number larger than $404$ can be attained by flipping the appropriate number of $-1$'s into $1's$. Then all even numbers $S$ which satisfy $404 \le |S| \le 2012$ can be obtained.