, 2015

There are an even number of solutions $(x_1, x_2, \dots, x_9)$ in which $x_1 \neq x_2$. Indeed they can be divided into pairs such that the two solutions in a pair are obtained from one another by swapping $x_1$ and $x_2$. Hence, so far as the parity of $N$ is concerned, one may assume $x_1 = x_2$. Likewise there are an even number of solutions satisfying $x_1 = x_2$ and $x_3 \neq x_4$; they can be divided into pairs such that the solutions in a pair are obtained from one another by swapping $x_3$ and $x_4$. So assume furthermore that $x_1 = x_2$ and $x_3 = x_4$. Solutions with this properties such that $x_1 \neq x_3$ can be divided into pairs again; the two solutions in a pair can be obtained from one another by swapping $x_1$ with $x_3$ and $x_2$ with $x_4$.

In summary we may restrict attention to solutions with $x_1 = x_2 = x_3 = x_4$. For them one can apply exactly the same reasoning to $x_5, x_6, x_7, x_8$; everything in the previous paragraph holds if the indices are increased by 4. So we need to determine the parity of the number of solutions of the form $(u, u, u, u, v, v, v, v, x_9)$. The ones among them with $u \neq v$ can be divided into pairs again, the solutions in a pair being $(u, u, u, u, v, v, v, v, x_9)$ and $(v, v, v, v, u, u, u, u, x_9)$.
Thus finally the question reduces to the parity of the number of solutions with $x_1 = x_2 = \dots = x_8$. In this case, denoting
$$x_1 = x_2 = \dots = x_8 = a,\ x_9 = b, \text{ we obtain the equation } \frac{8}{a} + \frac{1}{b} = 1.$$
Equivalently $8b+a=ab$, $(a-8)(b-1)=8$. Clearly $b \ge 2$, hence $b-1$ is a positive divisor of 8. The possibilities $b-1=1, 2, 4, 8$ yield respectively
$$a=16,\ b=2;\ a=12,\ b=3;\ a=10,\ b=5;\ a=9,\ b=9.$$
All 4 of these lead to solutions of the initial equation, and the solutions are distinct. Since $N$ has the parity of 4, it follows that it is even.