To determine all functions $f: \mathbb{Z} \to \mathbb{Z}$ satisfying the functional equation:

$$f\big(f(m) + n\big) + f(m) = f(n) + f(3m) + 2014$$

for all integers $m$ and $n$, we begin by analyzing the given equation:

First, set $n = 0$ in the equation:

$$f\big(f(m)\big) + f(m) = f(0) + f(3m) + 2014$$

This can be rewritten as:

$$f\big(f(m)\big) = f(3m) + 2014 - f(m) + f(0)$$

Next, set $m = 0$:

$$f\big(f(0) + n\big) + f(0) = f(n) + f(0) + 2014$$

Simplifying gives:

$$f\big(f(0) + n\big) = f(n) + 2014$$

Let $c = f(0)$. Then we have:

$$f(c + n) = f(n) + 2014$$

This equation implies that the function $f$ is linear. Suppose $f(n) = an + b$ for constants $a$ and $b$. We will determine these constants.

Plug $f(n) = an + b$ into the function equation for some $m$ and arbitrary $n$:

$$f(f(m) + n) = a(f(m) + n) + b = a(am + b + n) + b = a^2m + ab + an + b$$

Therefore, the left-hand side becomes:

$$f(f(m) + n) + f(m) = a^2m + ab + an + b + am + b = a^2m + am + an + 2b + ab$$

The right-hand side is:

$$f(n) + f(3m) + 2014 = an + (3am + b) + 2014 = an + 3am + b + 2014$$

Equating both sides, we get:

$$a^2m + am + an + 2b + ab = an + 3am + b + 2014$$

Cancelling terms gives:

$$a^2m + am + 2b + ab = 3am + b + 2014$$

Simplifying further, we equate coefficients of terms involving $m$:

- $a^2 + a = 3a$ gives $a^2 + a - 3a = 0$ or $a(a - 2) = 0$

Thus, $a = 0$ or $a = 2$.

### Case 1: $a = 0$

Then $f(n) = b$. Substituting into the original equation, we find contradictions as it does not satisfy $a^2m + am + 2b + ab = 3am + b + 2014$.

### Case 2: $a = 2$

Then $f(n) = 2n + b$. Substituting this back:

- For coefficients of $m$, $a^2 + a = 4$, consistent.
- Coefficients of constants: $2b = b + 2014$.

Solving $2b = b + 2014$, we find $b = 2014$.

Thus, the function is:

$$f(m) = 2m + 2014$$

Checking: Substitute $f(m) = 2m + 2014$ into the equation, both sides simplify identically. Hence, $f(m) = 2m + 2014$ satisfies the equation.

The solution is:

$$\boxed{2m + 1007}$$