Let $N$ be the number of ordered quintuples $\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right)$ of positive integers for which

$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}+\frac{1}{a_{5}}=1 .
$$

(For ordered quintuples, the order matters, so $(2,3,15,15,30)$ and $(15,2,15,3,30)$ are different ordered quintuples.)
Is $N$ an even or an odd number?
Consider an unordered quintuple that satisfies the condition and assume it consists of different numbers $b_{1}, \ldots, b_{k}$ (with $k \leq 5$), where $b_{i}$ occurs exactly $t_{i}$ times. Therefore, $t_{1}+\cdots+t_{k}=5$. This quintuple can be ordered in $\frac{5!}{t_{1}!\cdots t_{k}!}$ ways. This number is odd if and only if there are three factors of 2 in the denominator. This is true if and only if 4! or 5! appear in the denominator. In short, the only unordered quintuples that can be ordered in an odd number of ways are those consisting of at least four identical numbers.
Such a quintuple thus looks like ( $a, a, a, a, b$ ), where $b$ could possibly be equal to $a$ and where $a$ and $b$ are positive integers. We get the equation $\frac{4}{a}+\frac{1}{b}=1$, or equivalently $4 b+a=a b$. We see that $b$ must be a divisor of $a$ and that $a$ must be a divisor of $4 b$. Therefore, there are three options: $a=b$, $a=2 b$, and $a=4 b$. In the first case, we get $5 b=b^{2}$, so $b=5$. This gives the solution $(5,5,5,5,5)$. In the second case, we get $6 b=2 b^{2}$, so $b=3$. This gives the solution $(6,6,6,6,3)$. In the third case, we get $8 b=4 b^{2}$, so $b=2$. This gives the solution $(8,8,8,8,2)$. Therefore, there are three unordered quintuples that can be ordered in an odd number of ways. Thus $N$ is odd.