1.37 Four three-digit numbers with the same first digit are distinct and have the property: the sum of these numbers is divisible by any 3 of them. Find these 4 numbers.

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[Solution] Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the numbers we are looking for, $S$ be their sum, and $a$ be their common first digit. Thus, we have
$$100 a \leqslant x_{i} \leqslant 100(a+1), \quad (i=1,2,3,4)$$

Using these inequalities, we get
$$x_{i} + 300 a \leqslant S < x_{i} + 300(a+1)$$

From this, we have
$$1 + \frac{300 a}{x_{i}} \leqslant \frac{S}{x_{i}} < 1 + \frac{300(a+1)}{x_{i}}$$

Since $a \leqslant \frac{x_{i}}{100} < a+1$,
we have $\quad 1 + \frac{3 a}{a+1} < \frac{S}{x_{i}} < 4 + \frac{3}{a}$.
When $a=1$, we get $2.5 < \frac{S}{x_{i}} < 7$, and when $a \geqslant 2$, $3 < \frac{S}{x_{i}} < 5.5$.
Since three of the four quotients $\frac{S}{x_{i}}$ are integers and distinct, and there are only two integers between 3 and 5.5, the case $a \geqslant 2$ is impossible. Therefore, $a=1$ and the integer quotients should be among the numbers $3, 4, 5$, and 6.

Since the ratio of any two three-digit numbers between 100 and 199 is less than 2, 3 and 6 cannot both be quotients. Thus, only two possibilities remain: (1) the three integer quotients are $3, 4, 5$; (2) the three integer quotients are $6, 5, 4$.

In both cases, $S$ is divisible by 60. We set
$$S = 60 k.$$

In the first case, the numbers we are looking for are $12 k, 15 k, 20 k, 13 k$.
The first digit of all numbers is only 1 when $k=9$, so the four numbers we are looking for are $108, 135, 180, 117$.

In the second case, the numbers we are looking for are $10 k, 12 k, 15 k, 23 k$. Since the ratio of $23 k$ to $10 k$ is greater than 2, this case has no solution.

In conclusion, there is only one solution: $108, 135, 180, 117$.