Find an integral solution of the equation:  
\[ \left \lfloor \frac{x}{1!} \right \rfloor + \left \lfloor \frac{x}{2!} \right \rfloor + \left \lfloor \frac{x}{3!} \right \rfloor + \dots + \left \lfloor \frac{x}{10!} \right \rfloor = 2019. \]  

Note: \( \lfloor u \rfloor \) stands for the greatest integer less than or equal to \( u \).

Let's think step by step and output the final answer within \boxed{}.
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