7. (12 points) The sum of four different natural numbers is 2013, what is the minimum least common multiple of these four natural numbers?

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7. (12 points) The sum of four different natural numbers is 2013, what is the minimum least common multiple of these four natural numbers?
【Solution】Solution: Let the least common multiple be $x$, and the four numbers be $\frac{\mathrm{X}}{\mathrm{a}}, \frac{\mathrm{X}}{\mathrm{b}}, \frac{\mathrm{X}}{\mathrm{c}}$ and $\frac{\mathrm{X}}{\mathrm{d}}$. Then $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{x}}{\mathrm{b}}+\frac{\mathrm{x}}{\mathrm{c}}+\frac{\mathrm{x}}{\mathrm{d}}=\mathrm{x}\left(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}+\frac{1}{\mathrm{c}}+\frac{1}{\mathrm{~d}}\right)=2013=3 \times 11 \times 61$. To make $x$ as small as possible, $\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}+\frac{1}{\mathrm{c}}+\frac{1}{\mathrm{~d}}$ should be as large as possible. That is, $a, b, c, d$ should be as small as possible. Consider $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{25}{12}$. Then $\frac{25}{12} x=2013$, $x$ is not an integer, discard. Consider $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}=\frac{61}{30}$, $\frac{61}{30} \mathrm{x}=2013$, we get $x=990$. This satisfies the condition.
In conclusion, the least common multiple is 990. Therefore, the answer is: 990.