Find all integer solutions that satisfy the following conditions:
(1) $a, b, c$ are positive integers;
$$
\text{(2) } a+b+c=[a, b, c] \text{.}
$$
Solution: First, we discuss the case where $(a, b, c)=1$.

Assume $a \leqslant b \leqslant c$. Since $a+b+c=[a, b, c]$, then $c \mid (a+b)$. Therefore, we have $c=a+b$ or $2c=a+b$.

When $c=a+b$, since $2a+2b=[a, b, c]$, then $b \mid (2a+2b)$, which implies $b=a$ or $b=2a$.
- When $b=a$, $c=2a$. Given $(a, b, c)=1$, we get $a=1, b=1, c=2$. In this case, $a+b+c=[a, b, c]$ does not hold.
- When $b=2a$, $c=3a$. Given $(a, b, c)=1$, we get $a=1, b=2, c=3$. In this case, $a+b+c=[a, b, c]$ holds.

When $2c=a+b$, $a=b=c$. Given $(a, b, c)=1$, we get $a=1, b=1, c=1$. In this case, $a+b+c=[a, b, c]$ does not hold.

Therefore, when $(a, b, c)=1$, $a=1, b=2, c=3$.

If $(a, b, c)=n$, then $a+b+c=[a, b, c]$. Dividing both sides by $n$, we can reduce it to the previous case.

In summary, $a=n, b=2n, c=3n$, where $n$ is a positive integer.
(Fan Xingya, Gu Dan, Beijing No. 4 High School, 100034)