Will you divide the limit till this year?

$= \displaystyle \int_{0}^{tan1} 0 + \displaystyle \int_{tan1}^{2014} 1$

$= 2014 - tan1$

Note that $\tan^{-1} x$ is an increasing function for $x \in [0, 2014]$ and that $\tan^{-1} 2014 \approx 1.570$ . Then we have:

$\begin{aligned} I & = \int_0^{2014} \left \lfloor \tan^{-1} x \right \rfloor dx \\ & = \int_0^{\tan 1} \left \lfloor \tan^{-1} x \right \rfloor dx + \int_{\tan 1}^{2014} \left \lfloor \tan^{-1} x \right \rfloor dx \\ & = 0 + \int_{\tan 1}^{2014} dx \\ & = 2014 - \tan 1 \\ & = 2014 - \cot \left(\frac \pi 2 - 1\right) \end{aligned}$

Therefore, $\left \lfloor \dfrac ab \right \rfloor = \left \lfloor \dfrac {2014}{\frac \pi 2 - 1} \right \rfloor = \boxed{3528}$ .