Harvard-MIT Mathematics Tournament

Solution:

Let $\{x\} = x - \lfloor x\rfloor$ be the fractional part of $x$. Note that
$$\lfloor x\lfloor x\rfloor\rfloor = \lfloor (\lfloor x\rfloor + \{x\})\lfloor x\rfloor \rfloor = \lfloor x\rfloor^{2} + \lfloor \{x\}\lfloor x\rfloor \rfloor.$$
Because $\{x\}$ may take on any value in the half-open interval $[0,1)$, the quantity $\lfloor \{x\}\lfloor x\rfloor \rfloor$ can take on any integer value between $0$ and $\lfloor x\rfloor - 1$, inclusive.

If $\lfloor x\rfloor = n$, then $\lfloor x\lfloor x\rfloor\rfloor$ can be any of the numbers $n^{2}, n^{2} + 1, \ldots, n^{2} + n - 1$. In other words, there are precisely $n$ possible values that $\lfloor x\lfloor x\rfloor\rfloor$ can take, and moreover, all of them are less than $(n+1)^{2}$.

Because $44^{2} + 43 = 1979 < 2005$ and $45^{2} = 2025 > 2005$, $n$ can range between $1$ and $44$, inclusive. Therefore, the answer is
$$\sum_{n=1}^{44} n = \frac{44 \cdot 45}{2} = 990$$