Firstly, $$f\Big(\dfrac{1}{x}\Big)=\dfrac{1}{x^2+1} \implies f(x)= \dfrac{x^2}{x^2+1} \implies f(x) + f\Big(\dfrac{1}{x}\Big)=1$$ Therefore, we have \begin{align*} &f\Big(\dfrac{1}{2013}\Big)+f\Big(\dfrac{1}{2012}\Big)+f\Big(\dfrac{1}{2011}\Big)+\cdots +f\Big(\dfrac{1}{2}\Big)+f(1)+f(2)+\cdots +f(2011)+f(2012)+f(2013)\\ =&1\times 2012 + f(1)\\ =&2012 + \frac{1}{2}\\ =&\boxed{2012\frac{1}{2}} \end{align*}