Yanıt: $\boxed E$
$k$ negatif olmayan bir tam sayı olmak üzere;
$n = 3k$ ise $\left \lfloor \dfrac{2n}{3} \right \rfloor = \left \lfloor \dfrac{2 \cdot 3k}{3} \right \rfloor = 2k$
$n=3k+1$ ise $\left \lfloor \dfrac{2n}{3} \right \rfloor = \left \lfloor \dfrac{2 \cdot (3k+1)}{3} \right \rfloor = \left \lfloor 2k + \dfrac 23 \right \rfloor = 2k$
$n=3k+2$ ise $\left \lfloor \dfrac{2n}{3} \right \rfloor = \left \lfloor \dfrac{2 \cdot (3k+2)}{3} \right \rfloor = \left \lfloor 2k+1 + \dfrac 13 \right \rfloor = 2k+1$
$\begin{array}{lcl}
\displaystyle \sum_{n=1}^{100} \left \lfloor \dfrac{2n}{3} \right \rfloor &=& \displaystyle \sum_{k=1}^{33} 2k + \displaystyle \sum_{k=0}^{33} 2k + \displaystyle \sum_{k=0}^{32} (2k + 1) \\
&=& 2 \displaystyle \sum_{k=1}^{33} k + \displaystyle \sum_{k=1}^{66} k \\
&=& 2 \cdot \dfrac {33 \cdot 34}{2} + \dfrac {66\cdot 67}{2} \\
&=& 33 \cdot 34 + 33 \cdot 67 \\
&=& 33 \cdot 101 \\
&=& 3333
\end{array}$