For each integer $n \ge 1$, define $a_n = \left\lfloor \frac{n}{\left\lfloor \sqrt{n} \right\rfloor} \right\rfloor$, where $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$, for any real number $x$. Find the number of all $n$ in the set $\{1, 2, 3, \ldots, 2010\}$ for which $a_n > a_{n+1}$.

Let's think step by step and output the final answer within \boxed{}.
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