6. If $x$ is a real number, then the solution set of the equation
$$
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right]+\cdots+\left[\frac{x}{1993}\right]+
$$
$\left.〔 \frac{x}{1994}\right\rceil$ is $\qquad$ . (where $[x]$ denotes the greatest integer not exceeding $x$)
6. $\{0,4,5\}$.

Since $[a] \leqslant a3 .
$$
$\therefore$ The necessary condition for the above equation to hold is $0 \leqslant x<7$.
Note that the equation itself indicates that $x$ is an integer. Testing with $0,1,2,3,4,5$, 6, we find $x=0,4,5$.
$\therefore$ The solution set is $\{0,4,5\}$.