From strict to non-strict

We shall prove a more general inequality before tackling our specific problem: Given that $\frac{ab}{a + b} > n,$ where $n$ is a positive integer, we must have

$\frac{ab}{a + b} \geq n + \frac{1}{n^2 + 2n + 2}.$

First off, from the given condition $ab > n(a + b),$ we know that $ab = n(a + b) + r,$ where $r$ is some positive integer. This can be rewritten as $(a - n)(b - n) = n^2 + r$ via Simon's Favorite Factoring Trick. We must have $a, b > n,$ since if $a, b < n,$ then the upper bound for $(a - n)(b - n)$ is $n^2,$ which is less than $n^2 + r.$ Thus, we may write $a = a_1 + n, b = b_1 + n,$ where $a_1, b_1$ are positive integers.

This substitution causes our given condition to become $a_1b_1 = n^2 + r.$ Since $a_1, b_1 \geq 1,$ we have

$\begin{aligned} 0 &\leq (a_1 - 1)(b_1 - 1) \\ 0 &\leq a_1b_1 - a_1 - b_1 + 1 \\ a_1 + b_1 &\leq a_1b_1 + 1 \\ a_1 + b_1 &\leq n^2 + r + 1. \end{aligned}$

Also, we have $\frac{r}{x + r} \geq \frac{1}{x + 1}$ for all positive integers $x,$ since this reduces to $r \geq 1,$ which we know is true because $r$ is a positive integer.

Finally, incorporating our $a_1, b_1$ substitution into the expression $\frac{ab}{a + b}$ and applying the previous results yields

$\begin{aligned} \dfrac{ab}{a + b} &= \dfrac{n(a + b) + r}{a + b} \\ &= n + \dfrac{r}{a + b} \\ &= n + \dfrac{r}{2n + a_1 + b_1} \\ &\geq n + \dfrac{r}{2n + n^2 + r + 1} \\ &= n + \dfrac{r}{(n^2 + 2n + 1) + r} \\ &\geq n + \dfrac{1}{n^2 + 2n + 2}. \end{aligned}$

Equality holds when $r = 1$ and one of $a_1, b_1 = 1.$ This gives us the equality condition that $(a, b)$ is a permutation of $(n + 1, n^2 + n + 1). \, \, \blacksquare$

With regard to the problem, this inequality gives us $C = 2017 + \frac{1}{2017^2 + 2(2017) + 2}$ as the largest possible value of $C.$ The value of $q$ is clearly equal to $2017^2 + 2(2017) + 2,$ which has last three digits

$2017^2 + 2(2017) + 2 \equiv 17^2 + 2(17) + 2 \equiv \boxed{325} \! \pmod{1000}.$