$n \cdot \text{lcm}(a,b,2016) = ab-2016$ 
Find the largest even number $n$ such that there exist positive integers $a,b$ with $n \cdot \text{lcm}(a,b,2016) = ab-2016$.

I tried using the fact that $\text{lcm}(a,b,2016) \geq a,\text{lcm}(a,b,2016) \geq b,$ and $\text{lcm}(a,b,2016) \geq 2016$, but didn't see how to use this to solve the question. How should we approach it?

Clearly 2016 divides $ab$. Let $m=\mathrm{lcm}(a,b,2016)$; then
$m\ge2016$
$\implies\ ab-2016=nm\ge2016n$
$\implies\ n\le\dfrac{ab}{2016}-1$
Hence $n=\dfrac{ab}{2016}-1$ to be as large as possible. Then
$nm=ab-2016$ $\implies$ $m=2016$
So $a,b$ divide $m$ and $a,b\le m=2016$ $\implies$ $\dfrac{ab}{2016}\le2016$.
Also $n$ is even, so $\dfrac{ab}{2016}$ must be odd. The largest odd divisor of 2016 is 63. Hence we can take
$\dfrac{ab}{2016}=63$ (e.g. $a=2016,b=63$)
and so the largest even value of $n$ is
$$\boxed{n=62}$$
