1. Consider the general form of the equation:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{e} = 1$$
where $a, b, c, d, e \in \mathbb{Z}^{+}$.

2. **Count the solutions where $a \neq b$:**
If $(a, b, c, d, e)$ is a solution, then $(b, a, c, d, e)$ is also a solution. These pairs are distinct and thus the number of such solutions is even.

3. **Consider the case where $a = b$:**
The equation simplifies to:
$$\frac{2}{a} + \frac{1}{c} + \frac{1}{d} + \frac{1}{e} = 1$$
Similarly, if $(c, d, e)$ is a solution, then $(d, c, e)$ is also a solution, making the number of such solutions even.

4. **Consider the case where $a = b$ and $c = d$:**
The equation further simplifies to:
$$\frac{2}{a} + \frac{2}{c} + \frac{1}{e} = 1$$
Again, if $(a, c, e)$ is a solution, then $(c, a, e)$ is also a solution, making the number of such solutions even.

5. **Consider the case where $a = b$, $c = d$, and $a = c$:**
The equation now becomes:
$$\frac{4}{a} + \frac{1}{e} = 1$$
Solving for $e$, we get:
$$\frac{4}{a} = 1 - \frac{1}{e} \implies \frac{4e}{a} = e - 1 \implies 4e = ae - a \implies a(e - 1) = 4e \implies (a - 4)(e - 1) = 4$$

6. **Find the integer solutions to $(a - 4)(e - 1) = 4$:**
The integer pairs $(a - 4, e - 1)$ that satisfy this equation are:
$$(1, 4), (2, 2), (4, 1)$$
This gives us the solutions:
$$(a, e) = (5, 5), (6, 3), (8, 2)$$

7. Count the total number of solutions:
Since the number of solutions where $a \neq b$ and other similar cases are even, and we have exactly 3 solutions where $a = b = c = d$, the total number of solutions is odd.

$\blacksquare$