JEE quadratics 2

Observe one root is 2. Other root is -3/k obviously k should be negative .now,
the other root should lie between 2017 and 2018.we get k<-3/2018

$z=-k=\frac{a}{b}$ note that what ever comes next is on the assumption that $z>0$ ,otherwise we will have to multiply by -1 which will ruin the inequity. $\begin{aligned} -zx^2+2zx+3x-6>0\\ zx^2-2zx-3x+6<0\\ (zx-3)(x-2)<0\\ \end{aligned}$ so,from here we have 2 cases: $\begin{cases} (i)\rightarrow \frac{3}{z}<2\rightarrow \frac{3}{z}<x<2\\ (ii)\rightarrow \frac{3}{z}>2\rightarrow \frac{3}{z}>x>2 \end{cases}$ note that the integral values for n: $n_1<n<n_2$ ,is : $n_2-n_1-1$ .hence, $\begin{cases} \frac{3}{z}<x<2\rightarrow 2-\frac{3}{z}-1=2015\rightarrow z=-\frac{3}{2014}(rejected)\\ \frac{3}{z}>x>2\rightarrow \frac{3}{z}-2-1=2015\rightarrow z=\frac{3}{2018} \end{cases}$ this problem has nothing to do with maximum or minimum, a direct answer for z. $3+2018=\boxed{2021}$