, 2004

Solution:
We write the equation in the form:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{e} = \frac{1}{5}$$
The number of five tuples $(a, b, c, d, e)$ which satisfy the given relation and for which $a \neq b$ is even, because if $(a, b, c, d, e)$ is a solution, then so is $(b, a, c, d, e)$ which is distinct from $(a, b, c, d, e)$. Similarly, the number of five tuples which satisfy the equation and for which $c \neq d$ is also even. Hence it suffices to count only those five tuples $(a, b, c, d, e)$ for which $a = b$, $c = d$. Thus the equation reduces to
$$\frac{2}{a} + \frac{2}{c} + \frac{1}{e} = \frac{1}{5}$$
Here again, the tuple $(a, a, c, c, e)$ for which $a \neq c$ is even because we can associate a different solution $(c, c, a, a, e)$ to this five tuple. Thus it suffices to consider the equation
$$\frac{4}{a} + \frac{1}{e} = \frac{1}{5}$$
and show that the number of pairs $(a, e)$ satisfying this equation is odd.
This reduces to
$$a e = 20 e + 5 a$$
or
$$(a - 20)(e - 5) = 100$$
But observe that
$$\begin{aligned} & 100 = 1 \times 100 = 2 \times 50 = 4 \times 25 = 5 \times 20 \\ & \quad = 10 \times 10 = 20 \times 5 = 25 \times 4 = 50 \times 2 = 100 \times 1 \end{aligned}$$
Note that no factorisation of $100$ as product of two negative numbers yields a positive tuple $(a, e)$. Hence we get these $9$ solutions. This proves that the total number of five tuples $(a, b, c, d, e)$ satisfying the given equation is odd.