Argentine National Olympiad 2015

There are an even number of solutions $(x_1,x_2,\dots ,x_9)$ in which $x_1\ne x_2$. 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 $x_3\ne x_4$; 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 $x_1\ne x_3$ 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 $u\ne v$ 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 $x_1=x_2=\cdots =x_8$. In this case, denoting $$x_1=x_2=\cdots =x_8=a,x_9=b,\text{ we obtain the equation }\frac{8}{a}+\frac{1}{b}=1.$$ 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.