Yes, it exists.
There are finitely many fractions $\frac{a}{b}$ of lowest term such that $a \ge 1$, $1 \le b \le 2007$ and $\frac{a}{b} < \sqrt{2}$. Let $\frac{n}{m}$ be the largest such fraction. If $\lfloor \frac{nk}{m} \rfloor \ne \lfloor \sqrt{2}k \rfloor$ for some $k = 1, 2, \dots, 2007$, then $t = \lfloor \sqrt{2}k \rfloor$ is an integer satisfying $\frac{nk}{m} < t < \sqrt{2}k$ since $\frac{n}{m} < \sqrt{2}$ and $\sqrt{2}k \notin \mathbb{Z}$. It follows that $\frac{n}{m} < \frac{t}{k} < \sqrt{2}$, which contradicts the choice of $\frac{n}{m}$. Therefore, we must have $\lfloor \frac{nk}{m} \rfloor = \lfloor \sqrt{2}k \rfloor$ for $1 \le k \le 2007$.