Solution:
Obviously $n$ must be a positive integer. Now note that $44^2=1936<2014<2025=45^2$ and $12^3<1900<2014<13^3$.
If $n<1950$ then $2014=n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor<1950+44+12=2006$, a contradiction!
So $n \geq 1950$. Also if $n>2000$ then $2014=n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor>2000+44+12=2056$, a contradiction!
So $1950 \leq n \leq 2000$, therefore $\lfloor\sqrt{n}\rfloor=44$ and $\lfloor\sqrt[3]{n}\rfloor=12$. Plugging that into the original equation we get:
$$n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor=n+44+12=2014$$
From which we get $n=1958$.

Solution:
Obviously $n$ must be a positive integer. Since $n \leq 2014$, $\sqrt{n}<45$ and $\sqrt[3]{n}<13$.
From $n=2014-\lfloor\sqrt{n}\rfloor-\lfloor\sqrt[3]{n}\rfloor>2014-45-13=1956$, $\sqrt{n}>44$ and $\sqrt[3]{n}>12$, thus $\lfloor\sqrt{n}\rfloor=44$ and $\lfloor\sqrt[3]{n}\rfloor=12$ and $n=2014-\lfloor\sqrt{n}\rfloor-\lfloor\sqrt[3]{n}\rfloor=2014-44-12=1958$.