I think the bound can be made even stronger: $|f(x)-g(x)|>\pi$ My solution involved the use of $\pi\cdot\cot(\pi x)=\lim_{N\rightarrow\infty}\sum_{n=-N}^{N}\frac{1}{x+n}$, which was horrible.

Here's a solution for the original problem For each $j=0,1,2, \cdots ,2017$, define the interval $I_j=(j,j+1)$. Note that for every $0<\alpha <1$, $\frac{1}{\alpha(\alpha+1)}+\frac{1}{(1-\alpha)(2-\alpha)}=\frac{2n+2}{n(n+2)} >2$ since $0<n=\alpha-\alpha^2<\frac{1}{4}$ Now we seperate four cases : Case 1 : $x \in \bigcup_{j=1}^{1008} I_{2j-1}$

Case 2: $x \in \bigcup_{j=1}^{1008}I_{2j}$

Case 3 : $x \in I_{0}$

Case 4 : $x \in I_{2017}$

Hence, for all $x \in \bigcup_{j=0}^{2017}I_j$, $|f(x)-g(x)| >2$, as desired.

I got 3 points on this in competition because I ran out of time while solving, but I did finish it after getting home. After subtracting, $$f(x)-g(x)= \frac{1}{x}+\frac{1}{(x-1)(x-2)} + ... + \frac{1}{(x-2017)(x-2018)}$$Then, you divide it into cases of parity: you can determine that for $2n<x<2n+1$, the thing is huge and positive, and for $2n+1<x<2n+2$, it's huge and negative. For $2n<x<2n+1$, the case I proved in competition, it's somewhat easy: $\frac{1}{(x-2n+1)(x-2n)}+\frac{1}{(x-2n-1)(x-2n-2)}$ has minimum value when $x$ is right in the middle of $2n$ and $2n+1$, and just that in itself is bigger than 2, and as the rest of the sum is positive since all the terms are products of two negatives, the whole thing is bigger than 2. The other case is very similar but does require a bit more work.

Induction on $n\in \mathbb{Z}^+$ that: P(n) wrote: Let $f_n(x)=\sum_{i=0}^{n}{\frac{1}{x-2i}}$ and $g_n(x)=\sum_{i=1}^{n}{\frac{1}{x-(2i-1)}}$. Then $|f_n(x)-g_n(x)|>2$ for all non-integer $x$ that $0<x<2n$.

Achillys wrote: Let $f(x)$ and $g(x)$ be given by \[f(x) = \frac{1}{x} + \frac{1}{x-2} + \frac{1}{x-4} + \cdots + \frac{1}{x-2018}\]\[g(x) = \frac{1}{x-1} + \frac{1}{x-3} + \frac{1}{x-5} + \cdots + \frac{1}{x-2017}.\] Prove that $|f(x)-g(x)| >2$ for any non-integer real number $x$ satisfying $0 < x < 2018$. Solution. Define $h(x)=f(x)-g(x)=\sum_{i=0}^{2018}(-1)^i\frac{1}{x-i}.$ Note that $h(2018-x)=-h(x),$ so it suffices to check for the result for $x\in (0,1009).$ Claim 1. $-\frac{1}{x-(a-1)}+\frac{1}{x-a}+\frac{1}{(a+1)-x}-\frac1{(a+2)-x}>2$ for $x\in (a,a+1)$ where $a\in \mathbb N_0.$ Proof. Substitute $y=x-a,$ then $y\in(0,1)$ and we have to prove \[-\frac1{y+1}+\frac1{y}+\frac{1}{1-y}-\frac1{2-y}>2.\]Now, \begin{align*} -\frac1{y+1}+\frac1{y}+\frac{1}{1-y}-\frac1{2-y} &=\frac{1}{y(1-y)}-\frac3{(y+1)(2-y)}\\ &=\frac{2y-y^2+2-y-3y+3y^2}{y(1-y)(y+1)(2-y)}\\ &=\frac{2y^2-2y+2}{y(1-y)(y+1)(2-y)}\\ &=2\cdot\frac{y^2-y+1}{y^4-2y^3-y^2+2y} \end{align*}Hence, it suffices to prove $y^2-y+1>y^4-2y^3-y^2+2y$ that is \[y^4-2y^3-2y^2+3y-1<0\]Taking derivatives, \begin{align*} \left(y^4-2y^3-2y^2+3y-1\right)'=4y^3-6y^2-4y+3 &=4y^2\left(y-\frac12\right)-4y\left(y-\frac12\right)-6\left(y-\frac12\right)\\ &=2\left(y-\frac12\right)(2y^2-2y-3)\\ &=2\left(y-\frac12\right)\underbrace{\left(2\left(y-\frac12\right)^2-\frac72\right)}_{-\text{ve}} \end{align*}Therefore, $y^4-2y^3-2y^2+3y-1$ is increasing in $\left(0,\tfrac12\right]$ and decreasing in $\left[\tfrac12,1\right).$ Thus, \[y^4-2y^3-2y^2+3y-1<\frac1{16}-\frac14-\frac12+\frac32-1=-\frac3{16}-1<0.\]Hence, the claim. $\square$ Coming back to the original problem, Let $x\in(a,a+1)$ where $a\in \mathbb N_0.$ We now divide the proceedings into two cases that is whether $a$ is even or odd. Case (i). $a$ is even. \begin{align*} h(x)&=\frac1x-\frac{1}{x-1}+\frac1{x-2}-\frac{1}{x-3}+\cdots+\frac{1}{x-a}-\frac{1}{x-(a+1)}+\frac{1}{x-(a+2)}-\cdots+\frac1{x-2018}\\ &=\frac1x+\underbrace{\left(-\frac{1}{x-1}+\frac1{x-2}\right)}_{+\text{ve}}+\underbrace{\left(-\frac{1}{x-3}+\frac1{x-4}\right)}_{+\text{ve}}+\cdots+\frac{1}{x-a}+\frac{1}{(a+1)-x}-\frac{1}{(a+2)-x}+\frac{1}{(a+3)-x}-\cdots-\frac1{2018-x}\\ &>\frac1x-\frac{1}{x-(a-1)}+\frac{1}{x-a}+\frac1{(a+1)-x}-\frac{1}{(a+2)-x}+\underbrace{\left(\frac{1}{(a+3)-x}-\frac{1}{(a+4)-x}\right)}_{+\text{ve}}+\cdots+\underbrace{\left(\frac{1}{2017-x}-\frac{1}{2018-x}\right)}_{+\text{ve}}\\ &>-\frac{1}{x-(a-1)}+\frac{1}{x-a}+\frac1{(a+1)-x}-\frac{1}{(a+2)-x}\\ &>2 \end{align*}Where the last step follows from Claim 1. Case (ii). $a$ is odd. \begin{align*} h(x)&=\frac1x-\frac{1}{x-1}+\frac1{x-2}-\frac{1}{x-3}+\cdots-\frac{1}{x-a}+\frac{1}{x-(a+1)}-\frac{1}{x-(a+2)}-\cdots+\frac1{x-2018}\\ &=\underbrace{\left(-\frac{1}{x}+\frac1{x-1}\right)}_{-\text{ve}}+\underbrace{\left(-\frac{1}{x-2}+\frac1{x-3}\right)}_{-\text{ve}}+\cdots-\frac{1}{x-a}-\frac{1}{(a+1)-x}+\frac{1}{(a+2)-x}-\frac{1}{(a+3)-x}-\cdots-\frac1{2018-x}\\ &<\frac1{x-(a-1)}-\frac1{x-a}-\frac{1}{(a+1)-x}+\frac1{(a+2)-x}+\underbrace{\left(-\frac{1}{(a+3)-x}+\frac{1}{(a+4)-x}\right)}_{-\text{ve}}+\cdots\underbrace{\left(-\frac{1}{2017-x}+\frac{1}{2018-x}\right)}_{-\text{ve}}\\ &<\frac1{x-(a-1)}-\frac1{x-a}-\frac{1}{(a+1)-x}+\frac1{(a+2)-x}\\ &=-\left(-\frac1{x-(a-1)}+\frac1{x-a}+\frac{1}{(a+1)+x}-\frac1{(a+2)-x}\right)\\ &<-2 \end{align*}Again, the last step follows from Claim 1. Thus, we obtain $h(x)>2$ when $\lfloor x\rfloor$ is even and otherwise $h(x)<-2.$ And we are done. $\blacksquare$

Is there a solution along the lines of like finding the minimum of $\frac1{x(1-x)} $ on the interval $(0, 1)$ to be $-4$, and doing casework and using the dumb bound of the sum of squares is $\frac{\pi^2}6$?

@above lol I'm pretty sure someone I knew used that. The issue is that like square bounds are not the "best" but the product between consecutive ones are the best. Personally, I used the fact that $ln(2)=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...$

Let $\lfloor{x}\rfloor=m$, we divide into two cases: Case I: $m$ is odd, let $m=2k+1$, then we have \begin{align*} &\frac{1}{x-m}+\frac{1}{m+1-x}-\left(\frac{1}{x-m+1}+\frac{1}{m+2-x}\right)\\ &=\frac{1}{(x-m)(m+1-x)}-\frac{3}{(x-m+1)(m+2-x)}\\ &=\frac{(x-m+1)(m+2-x)-3(x-m)(m+1-x)}{(x-m)(m+1-x)(x-m+1)(m+2-x)}\\ &=\frac{2x^2-(4m+4)x+2m^2+2m+2}{(x-m)(m+1-x)(x-m+1)(m+2-x)} \end{align*}Notice that this expression is minimized when $x=\frac{2m+1}{2}$, and when $x=\frac{2m+1}{2}$ this expression equals $4-\frac{2}{3}>2$. On the other hand we have $$\frac{1}{x-2a-1}\geq \frac{1}{x-2a}$$for every $0\leq a\leq k-1$ and $$\frac{1}{2a-x}\geq\frac{1}{2a+1-x}$$for every $k+2\leq a\leq 1008$ and $$\frac{1}{2018-x}>0$$summing them yields $g(x)-f(x)>2$. Case II: $m$ is even, let $m=2k$ Similarly we have $$\frac{1}{x-m}+\frac{1}{m+1-x}-\left(\frac{1}{x-m+1}+\frac{1}{m+2-x}\right)>2$$$$\frac{1}{x}>0$$$$\frac{1}{x-2a}>\frac{1}{x-2a+1}$$for each $1\leq a\leq k-1$ and $$\frac{1}{2a-1-x}>\frac{1}{2a-x}$$for each $k+1\leq a\leq 1009$ summing them yields $f(x)-g(x)>2$.

lminsl wrote: Here's a solution for the original problem For each $j=0,1,2, \cdots ,2017$, define the interval $I_j=(j,j+1)$. Note that for every $0<\alpha <1$, $\frac{1}{\alpha(\alpha+1)}+\frac{1}{(1-\alpha)(2-\alpha)}=\frac{-2n+2}{n(n+2)} \geq 8/3$ since $0<n=\alpha-\alpha^2\leq\frac{1}{4}$ Now we seperate four cases : Case 1 : $x \in \bigcup_{j=1}^{1008} I_{2j-1}$

Case 2: $x \in \bigcup_{j=1}^{1008}I_{2j}$

Case 3 : $x \in I_{0}$

Case 4 : $x \in I_{2017}$

Hence, for all $x \in \bigcup_{j=0}^{2017}I_j$, $|f(x)-g(x)| \geq8/3 \textgreater 2$, as desired.

Official solution 1 is pretty nice. It does some optimizations first, which gives that it suffices to solve the problem for the case $$ x \in (1,2) $$and working of this case is much easier, and way more intuitive to human mind. Otherwise, we have to do a messy case bash.

Disgusting bounding. Basically we can look at the specific interval it lies in, bound with alternating series theorem and with some manipulation of $\frac{\pi^2}{6}$ for $\frac{1}{6}+\frac{1}{20} + \ldots $ to get a working bound to hit $2$. My thing eventually reduced to $\pi^2 < \frac{253}{24}$ which is true since $\pi^2 < 10$.