Four different three-digit numbers start with the same digit. Their sum is divisible by three of the numbers. Determine the numbers.

 untranslated portion:
Négy egymástól különböző, háromjegyű szám ugyanazzal a számjeggyel kezdődik. Összegük osztható a számok közül hárommal. Határozzuk meg a számokat. 

Translated portion:
Four different three-digit numbers start with the same digit. Their sum is divisible by three of the numbers. Determine the numbers.
Let the four numbers be $a, b, c, d$, their common leading (hundreds place) digit be $j (\geqq 1)$, their sum be $s$, and further, $s / a=\alpha$, $s / b=\beta, s / c=\gamma$ are natural numbers, and the size relationship of the first three numbers (which are divisors of $s$) is:

$$
100 j \leq a\beta>\gamma$, and

$$
1c+(100 j+100+0)= \\
=c+100(j+1)>2 c, \quad \gamma>2
\end{gathered}
$$

According to these, the values of $\alpha, \beta$, $\gamma$ must be among 6, 5, 4, and 3, but due to (2), 6 and 3 cannot appear together. Thus, their values are either 6, 5, and 4, or 5, 4, and 3.

In the first case,

$$
\begin{aligned}
a+b+c & =s\left(\frac{1}{6}+\frac{1}{5}+\frac{1}{4}\right)=\frac{37 s}{60} \\
d & =\frac{23 s}{60}>\frac{s}{3}=2 a
\end{aligned}
$$

which, according to (1a), is impossible.

In the second case, similarly,

$$
\begin{gathered}
a+b+c=s\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)=\frac{12 s}{60}+\frac{15 s}{60}+\frac{20 s}{60}=\frac{47 s}{60} \\
d=\frac{13 s}{60}
\end{gathered}
$$

thus, the largest number among our numbers is $c$, and the smallest is $a$. Furthermore, considering that $d$ is an integer, $s / 60=k$ is an integer.

Due to the common leading digit,

$$
\begin{gathered}
c-a=20 k-12 k=8 k400 j$, thus $j<2$, so $j=1$. Now, with $c=20 k<200$, we have $k \leq 9$, and with $a=12 k \geq 100$, we have $k \geq 9$, so $k=9, s=540, a=108, b=135, c=180, d=117$.