• Prove that $A_k \neq 0$ and $B_k \neq 0$ for all $k$
• Prove that $A_k+B_k \neq 2017m$ for all $k$ and any integer $m$
•  For $x,y$ with $x+y = 2017$, prove that $A_x+A_y=B_x+B_y=2017$
• For $x,y$ with $x+y = 2017$, prove that, either $A_x+B_x>2017$ and $A_y+B_y<2017$

Step 1:

Prove that $A_k \neq 0$ and $B_k \neq 0$ for all $k$

Proof: For the sake of contradiction, assume that $A_k = 0$.

But, $0 = A_k \equiv ak \pmod{2017}$.

So, $2017\mid ak$

But, 2017 is a prime. So, $2017\mid a$ or $2017\mid k$

But, $a,k<2017$. So, $2017\nmid a$ and $2017\nmid k$

Contradiction. In the same way, $B_k\neq 0$ can be proved.

Step 2: 

Prove that $A_k+B_k \neq 2017m$ for all $k$ and any integer $m$

Proof: For the sake of contradiction, assume that $A_k + B_k \equiv 0 \pmod{2017}$.

But, $0 \equiv A_k+B_k \equiv ak+bk \equiv (a+b)k \pmod{2017}$.

So, $2017\mid (a+b)k$

But, 2017 is a prime. So, $2017\mid a+b$ or $2017\mid k$

But, $a,b,k<2017$ and $a+b\neq 2017$. So, $2017\nmid a+b$ and $2017\nmid k$

Contradiction.

Step 3: 

For $x,y$ with $x+y = 2017$, prove that $A_x+A_y=B_x+B_y=2017$

$A_x+A_y\equiv ax+ay \equiv a(x+y) \equiv 0 \pmod{2017}$

But, $A_x,A_y \neq 0$ and $A_x,A_y<2017$ (remainder when divided by 2017).

So, $0< A_x+A_y < 2\cdot 2017$

So, $A_x+A_y = 2017$

In the same way, $B_x+B_y=2017$ can be proved.

Step 4: 

For $x,y$ with $x+y = 2017$, prove that,  either $A_x+B_x>2017$ and $A_y+B_y<2017$ 

or $A_y+B_y>2017$ and $A_x+B_x<2017$

Proof: From step 3, we have $A_x+A_y=B_x+B_y=2017$

So, $(A_x+B_x)+(A_y+B_y) = A_x+A_y+B_x+B_y = 2\cdot 2017$

But, $A_x+B_x \neq 2017$ from step 2.

So, either $A_x+B_x > 2017 \implies A_y+B_y < 2\cdot2017-2017 = 2017$

Or, $A_x+B_x < 2017 \implies A_y+B_y > 2\cdot2017-2017 = 2017$.

So, with $x+y=2017$, exactly one of $A_x+B_x$ and $A_y+B_y$ is greater than 2017.

There are $\frac{2016}{2} = 1008$ such pairs $(x,y)$. So, there are exactly 1008 $k$ with $A_k+B_k>2017$.