1. Understanding Zero-Requiem Functions:
A function $\psi: \mathbb{Z} \to \mathbb{Z}$ is zero-requiem if for any positive integer $n$ and any integers $a_1, a_2, \ldots, a_n$, the sums $a_1 + a_2 + \cdots + a_n$ and $\psi(a_1) + \psi(a_2) + \cdots + \psi(a_n)$ are not both zero.

2. Given Conditions:
- $f$ and $g$ are zero-requiem functions.
- $f \circ g$ and $g \circ f$ are the identity functions, meaning $f$ and $g$ are mutually inverse bijections.
- $f + g$ is not a zero-requiem function.

3. Objective:
Prove that $f \circ f$ and $g \circ g$ are both zero-requiem functions.

4. Lemma:
Suppose $f$ is a zero-requiem function. Then for all $a_1 + \cdots + a_n = 0$, $f(a_1) + \cdots + f(a_n)$ have the same sign.

Proof:
Suppose $a_1 + \cdots + a_n = 0$ and $b_1 + \cdots + b_n = 0$. Assume $f(a_1) + \cdots + f(a_n) > 0$ and $f(b_1) + \cdots + f(b_n) < 0$. Let $A = f(a_1) + \cdots + f(a_n)$ and $B = -(f(b_1) + \cdots + f(b_n))$. Then:
$$B(a_1 + a_2 + \cdots + a_n) + A(b_1 + \cdots + b_n) = 0$$
$$B(f(a_1) + \cdots + f(a_n)) + A(f(b_1) + \cdots + f(b_n)) = 0$$
This leads to a contradiction since $A > 0$ and $B > 0$. Hence, $f(a_1) + \cdots + f(a_n)$ must have the same sign for all $a_1 + \cdots + a_n = 0$.

5. **Behavior of $f$ and $f^{-1}$**:
Since $f$ is zero-requiem, without loss of generality, assume for all $a_1 + \cdots + a_n = 0$, $f(a_1) + \cdots + f(a_n) > 0$. If for all $b_1 + \cdots + b_n = 0$, $f^{-1}(b_1) + \cdots + f^{-1}(b_n) > 0$, then $f + f^{-1}$ would be zero-requiem, which contradicts the given condition. Therefore, for all $b_1 + \cdots + b_n = 0$, $f^{-1}(b_1) + \cdots + f^{-1}(b_n) < 0$.

6. **Sign Behavior of $f$**:
Either $f(x) > 0$ for all $x > 0$ or $f(x) > 0$ for all $x < 0$.

Proof:
Suppose $f(a) \leq 0$ and $f(-b) \leq 0$ where $a, b > 0$. Then:
$$\underbrace{f(a) + f(a) + \cdots + f(a)}_{b} + \underbrace{f(b) + f(b) + \cdots + f(b)}_{a} = bf(a) + af(b) < 0$$
This inequality holds since $f$ is a bijection.

7. Case Analysis:
- Case 1: $f(x) > 0$ for all $x > 0$.
- This implies $f^{-1}(x) < 0$ for all $x < 0$.
- Let $m = \inf_{x > 0} \frac{f(x)}{x}$ and $n = \sup_{x < 0, f(x) < 0} \frac{f(x)}{x}$.
- $m > 0$ and $n > 0$.
- If $m < n$, there exist $a > 0$ and $b > 0$ such that:
$$f(a) \leq a(m + \frac{n - m}{3}) \quad \text{and} \quad f(-b) \leq -b(n - \frac{n - m}{3})$$
Thus, $af(-b) + bf(a) < 0$, which is absurd.

- Therefore, $m \geq n$. For $a_1 + \cdots + a_k = b_1 + \cdots + b_l$ where $a_i, b_i \geq 0$:
$$f(f(a_i)) \geq m^2 a_i \quad \text{and} \quad f(f(-b_i)) \geq -n^2 b_i$$
Summing gives:
$$\sum f(f(a_i)) + \sum f(f(-b_j)) \geq m^2 \sum a_i - n^2 \sum b_j = (\sum a_i)(m^2 - n^2)$$
This is zero only if $m = n$ and all equalities hold, implying $f$ is not zero-requiem. Hence, in this case, $f^2$ is zero-requiem.

- Case 2: $f(x) > 0$ for all $x < 0$.
- This implies $f^{-1}(y) < 0$ for all $y > 0$.
- Since $f^{-1}(x) < 0$ for all $x > 0$ or $f^{-1}(x) < 0$ for all $x < 0$, the latter is impossible because it contradicts the previous implication.
- Hence, $f^{-1}$ and $f$ map $\mathbb{N}^+$ to $\mathbb{Z}^-$ and $\mathbb{Z}^-$ to $\mathbb{N}^+$, implying $f(0) = 0$, which is absurd.

8. Conclusion:
Both $f \circ f$ and $g \circ g$ are zero-requiem functions.

$\blacksquare$