The answer is $652400$. Notice that $y\lfloor x \rfloor + z\lfloor y \rfloor + x\lfloor z \rfloor \leq xy + yz + zx$. On the other hand, $xy + yz + zx \leq \frac{1}{3}(x + y + z)^2$. It follows that
$$y\lfloor x \rfloor + z\lfloor y \rfloor + x\lfloor z \rfloor \leq \frac{1}{3}(x + y + z)^2 = 652400 + \frac{1}{3}.$$
If $x$, $y$, $z$ are integers then $y\lfloor x \rfloor + z\lfloor y \rfloor + x\lfloor z \rfloor \leq 652400$. If at least one of them is not integer it follows from the fact that $\{x\} + \{y\} + \{z\} \in \mathbb{Z}$ that $\{x\} + \{y\} + \{z\} \in \{1, 2\}$. Hence, at least one of them has a fractional part that is at least $\frac{1}{3}$. Yielding
$$y\lfloor x \rfloor + z\lfloor y \rfloor + x\lfloor z \rfloor \leq 652400 + \frac{1}{3} - \frac{1}{3} = 652400.$$
The equality case occurs at $(x, y, z) = (466, 466, 467)$, for example. ■