Hope I didn't make some mistake again... First of all, put origin of cartesian coordinate system in lower left corner. There are four types of these lines: horizontal, vertical, with positive slope, and with negative slope. There are $ 180n + 8$ horizontal and vertical lines, so its divisible with 4. Number of lines with positive slope is the same as the number of lines with negative slope, by simetry, so the number of all lines is divisible with 4 iff number of lines with positive slope is even. We are going to prove that: The slopes of these lines are rational numbers of the form: $ \frac ba, a\in {1,...,90n + 1\}, b\in \{1,...,90n + 5\}, gcd(a,b)=1}$, and let $ A$ be the set of all the slopes of these lines (there is no line with slope smaller then $ \frac 1{90n + 1}$ or bigger then $ 90n + 5$). It's not hard to se that number of lines with slope $ \frac ba$ is $ b\cdot (90n + 1 + 1 -a) + a\cdot (90n + 5 + 1 - b) - ab = b(90n + 2 ) + a(90n + 6) - 3ab\equiv_2 ab$. If $ \frac ba \in A$ and $ (a,b)\not = (1,1)$ and $ b\leq 90n + 1$ then $ \frac ab$ is also in $ A$, so the number of lines with slope $ \frac ba$ + the number of lines with slope $ \frac ab$ is $ 2ab \mod 2$ so it's even. Now we have to count lines with slopes $ 1, \frac {90n + 2}a, \frac {90n + 3}a, \frac {90n + 4}a, \frac {90n + 5}a$. There are $ 180n + 5$ lines with slope 1, and there are even number of lines with slopes of the form $ \frac {90n + 2}a$ and $ \frac {90n + 4}a$, so we have to deal with slopes $ \frac {90n + 3}a$ and $ \frac {90n + 5}a$. We saw that if $ a$ is even, then number of lines with slope $ \frac {90n + 3}a$ is even, but when $ a$ is odd, then there is odd number of those lines. But there are even number of $ a$'s which are relatively prime with $ 90n + 3$ and odd ($ 4|\varphi(90n + 3)$), so the number of all lines with slope $ \frac {90n + 3}a$ is even. On the other hand, there is odd number of odd $ a$ which are relatively prime with $ 90n + 5$ and not bigger then $ 90n + 1$ (again $ 4|\varphi (90n + 5)$ but $ a$ can't be $ 90n+3$) so there is odd number of lines with slopes of the form $ \frac {90n + 5}a$, but the number of lines with slope $ 1$ is also odd. So, the number of all lines with positive slope is even. QED

hmmm. can't you just take the two perpendicular bisectors of the sides of the rectangle as your x and y axis? then for each such line except the lines passing through the origin and the lines parallel to the axis there will be 4 images (reflection on x axis, y axis 180 degrees rotation by the origin and the line itslef) that are distinct and this gives a partition into four-tuples for these lines. together with the 180n+8 horizontal or vertical lines we only need to show that the number of lines passing through the origin is divisible by 4. now this number corresponds to twice the number of rational numbers $ \frac{a}{b}$ with $ a\in \{1,3,5,\dots, 90n+1\},b\in \{1,3,\dots,90n+5\}$ which we have to show is even. as we can pair $ \frac{a}{b}$ with $ \frac{b}{a}$ then we only need to consider $ b\in \{90n+3,90n+5\}$ which gives us $ \varphi (90n+5)+\varphi (90n+3)$ the first term is divisible by 4 as $ 5|90n+5$ and the second term is divisible by four as $ 90n+3$ contains at least two odd prime divisors

Well, yeah, yours is a bit smarter When I solved it, i didn't try to improve it. I am just not sure, how do you exclude cases $ b\in \{90n + 2, 90n + 4\}$?

note that in my coordinate system the points can not have integer coordinates, instead they are of the form $ (\frac{2p+1}{2},\frac{2q+1}{2})$ do you see it now?

yuup bye

dude no guys this method fails. here's a hint: are you sure (Albanian Eagles) that what you want to consider is phi(90n+5)+phi(90n+3) ?

Yes pretty sure.

no dude listen you want to consider gamma(93)+gamma(95), where the gamma function is the number of ODD numbers relatively prime to whatever. btw if you use your method you end up with 2 mod 4, since the line y=x is paired with y=x when you pair reciprocal slopes. that's why you actually need to show gamma(93)+gamma(95) is odd rather than phi(93)+phi(95) is even

another way:(Hope i don't mistake .please check it) you can take perpendicular bisector of the sides of rectangle the $X,Y$ axes.the number of vertical and horizontal lines are $180n+8$ that is divisible by $4$.the number of lines that pass the origin is $2(45n+3)+2(45n+1)$ that is divisible by $4$.rest of the lines can be partitioned into groups such that the number of each group is $4$.because we can consider of reflections of each line through the axes$X,Y$. and that is exactly $4$.