Note that for terms $i \in [1,3], [\sqrt i ] = 1$ $i \in [4,8], [\sqrt i ] = 2$ $i \in [9,15], [\sqrt i ] = 3$ So, $[\sqrt i ] = n$ for $2n+1$ integers of $n$. Thus we can express the sum as $\sum_{k=1}^{43} (n(2n+1)) + \sum_{k=44^2}^{2001} 44$ $2 \sum n^2 + \sum n + 66 \cdot 44$ $\dfrac{43 \cdot 44 \cdot 87}{3} + \dfrac{43 \cdot 44}{2} + 66 \cdot 44$ $\boxed{58718}$

I have an another approach.We can consider on cartesian coordinates.The value $\sum_{i=1}^{2001} [\sqrt i]$ denotes the number of lattice point $(x>0,y>0)$ below $y=\sqrt x$.We can calculate the number of lattice point by considering the number of lattice point $(y>0,x>0)$ below $x=y^2$.Namely, $44\cdot 2001-\sum_{y=1}^{44} (y^2-1)=\boxed{58718}$.