4. The set $\{[x]+[2 x]+[3 x] \mid x \in \mathbf{R}\} \bigcap\{1,2, \cdots, 100\}$ has $\qquad$ elements, where $[x]$ denotes the greatest integer not exceeding $x$.


Let $x=k+\alpha$, when $\alpha \in\left[0, \frac{1}{3}\right)$, $[x]+[2 x]+[3 x]=6 k$; 
when $\alpha \in\left[\frac{1}{3}, \frac{1}{2}\right)$, $[x]+[2 x]+[3 x]=6 k+1$; when $\alpha \in\left[\frac{1}{2}, \frac{2}{3}\right)$, $[x]+[2 x]+[3 x]=6 k+2$; $\alpha \in\left[\frac{2}{3}, 1\right)$, $[x]+[2 x]+[3 x]=6 k+3$. Therefore, there are $3+16 \cdot 4=67$ elements.