Let $ \{x\}=x-[x]$. Clearly, $ 0 \leq \{x\} < 1$. We have three cases: a) If $ 0 \leq \{x\} < \frac{1}{3}$, then $ [x]=\left[x+\frac{1}{2}\right]=\left[x+\frac{2}{3}\right]=r \rightarrow 3r=2002 \rightarrow r$ is not an integer, contradiction. b) If $ \frac{1}{3} \leq \{x\} < \frac{1}{2}$, then $ [x]=\left[x+\frac{1}{2}\right]=r$ and $ \left[x+\frac{2}{3}\right]=r+1$, so $ 3r+1=2002 \rightarrow r=667$. Hence all $ x \in \mathbb R$ such that $ 667+\frac{1}{3} \leq x <667+\frac{1}{2}$ satisfy the conditions. c) If $ \frac{1}{2} \leq \{x\} < 1$, then $ [x]=r$ and $ \left[x+\frac{1}{2}\right]=\left[x+\frac{2}{3}\right]=r+1$, hence $ 3r+2=2002 \rightarrow r$ is not an integer, contradiction. So the only solutions are the ones we found on b).

See Hermite's identity.