4. Find the integer solution to the equation $\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{10!}\right]=1001$.
4. From the equation, it can be directly determined that $x$ is a positive integer not exceeding 1001, hence $x<6$!, so the last five terms on the left side of the equation can be omitted. Each positive integer $x<6$! can be uniquely represented as
$$
x=a \cdot 5!+b \cdot 4!+c \cdot 3!+d \cdot 2!+e \cdot 1!
$$

where $a, b, c, d, e$ are non-negative integers, and $a \leqslant 5, b \leqslant 4, c \leqslant 3, d \leqslant 2, e \leqslant 1$. Substituting (3) into the original equation gives $206 a+41 b+10 c+3 d+e=1001$.

Since $41 b+10 c+3 d+e \leqslant 201$, it follows that $800 \leqslant 206 a \leqslant 1001$, from which we get $a=4$, thus $41 b+10 c+3 d+e=177$. Similarly, we can determine $b=4, c=1, d=1, e=0$, so $x=$ $4 \times 5!+4 \times 4!+3!+2!=584$ is the positive integer solution to the equation.