Answer: C

Solution:

The general term is $\binom{1000}{k}(x\sqrt[3]{2})^{1000-k}(y\sqrt{3})^{k},$ whose coefficient contains $2^{(1000-k)/3}$ and $3^{k/2}.$ This is rational exactly when $3 \mid (1000 - k)$ and $k$ is even.

Since $1000 \equiv 1 \pmod 3,$ we need $k \equiv 1 \pmod 3$ and $k$ even, which combine to $k \equiv 4 \pmod 6.$ The valid values $k = 4, 10, \ldots, 1000$ number $\dfrac{1000 - 4}{6} + 1 = 167.$

Thus, the correct answer is C.