4. Let $[x]$ denote the greatest integer not exceeding $x$. Then the set $\{[x]+[2 x]+[3 x] \mid x \in R\} \mid\{1,2, \ldots, 100\}$ has
$\qquad$ elements.
4. 67

Analysis: Let $f(x)=[x]+[2 x]+[3 x]$, then $f(x+1)=f(x)+6$, so we only need to consider the range of values on $[0,1)$.

When $x \in\left[0, \frac{1}{3}\right)$, $f(x)=0$; when $x \in\left[\frac{1}{3}, \frac{1}{2}\right)$, $f(x)=1$; when $x \in\left[\frac{1}{2}, \frac{2}{3}\right)$, $f(x)=2$; when $x \in\left[\frac{2}{3}, 1\right)$, $f(x)=3$; in summary, the range of $f(x)$ on $[0,1)$ is $\{0,1,2,3\}$. Since $100=16 \times 6+4$, there are 67 elements in the intersection.