Suppose $(a_n)$, $(b_n)$, $(c_n)$ are arithmetic progressions. Given that $a_1+b_1+c_1 = 0$ and $a_2+b_2+c_2 = 1$, compute $a_{2014}+b_{2014}+c_{2014}$.

[i]Proposed by Evan Chen[/i]
1. Given that \( (a_n) \), \( (b_n) \), and \( (c_n) \) are arithmetic progressions, we can express them as follows:
   \[
   a_n = a_1 + (n-1)d_a, \quad b_n = b_1 + (n-1)d_b, \quad c_n = c_1 + (n-1)d_c
   \]
   where \( d_a \), \( d_b \), and \( d_c \) are the common differences of the sequences \( (a_n) \), \( (b_n) \), and \( (c_n) \) respectively.

2. We are given the initial conditions:
   \[
   a_1 + b_1 + c_1 = 0
   \]
   and
   \[
   a_2 + b_2 + c_2 = 1
   \]

3. Using the general form of the arithmetic progressions, we can write:
   \[
   a_2 = a_1 + d_a, \quad b_2 = b_1 + d_b, \quad c_2 = c_1 + d_c
   \]

4. Substituting these into the second given condition:
   \[
   (a_1 + d_a) + (b_1 + d_b) + (c_1 + d_c) = 1
   \]
   Simplifying, we get:
   \[
   a_1 + b_1 + c_1 + d_a + d_b + d_c = 1
   \]

5. Using the first given condition \( a_1 + b_1 + c_1 = 0 \), we substitute:
   \[
   0 + d_a + d_b + d_c = 1
   \]
   Therefore:
   \[
   d_a + d_b + d_c = 1
   \]

6. Let \( d_n = a_n + b_n + c_n \). Since \( d_n \) is also an arithmetic progression, we can express it as:
   \[
   d_n = d_1 + (n-1)d_d
   \]
   where \( d_d \) is the common difference of \( d_n \).

7. From the given conditions, we know:
   \[
   d_1 = a_1 + b_1 + c_1 = 0
   \]
   and
   \[
   d_2 = a_2 + b_2 + c_2 = 1
   \]

8. The common difference \( d_d \) of \( d_n \) can be found as:
   \[
   d_d = d_2 - d_1 = 1 - 0 = 1
   \]

9. Therefore, the general form of \( d_n \) is:
   \[
   d_n = 0 + (n-1) \cdot 1 = n - 1
   \]

10. To find \( d_{2014} \):
    \[
    d_{2014} = 2014 - 1 = 2013
    \]

The final answer is \(\boxed{2013}\).