Solution:

The only solution triples are $(1,1,1)$, $(1,1,2)$ and $(1,2,3)$.

Let us first show that $a, b, c$ are pairwise coprime (we only show that $\operatorname{MCD}(a, b)=1$; for the other pairs the proof is the same).
If $d$ is the greatest common divisor of $a$ and $b$, then $d$ divides $a$, which in turn divides $b+c$, so $d$ divides $b+c$; but $d$ divides $b$, so it also divides $b+c-b=c$.
Then $d$ is simultaneously a divisor of $a, b$ and of $c$, and hence $d=1$, since $\operatorname{MCD}(a, b, c)=1$ by hypothesis.

We now note that $a$ divides $b+c$ by hypothesis, so $a$ also divides the sum $(b+c)+a$; similarly we obtain that $b$ and $c$ also divide $a+b+c$.
Since $a, b, c$ are pairwise coprime, the fact that each of them divides the sum $a+b+c$ implies that the product $a b c$ also divides $a+b+c$.
All divisors of a natural number are less than or equal to the number itself, so a necessary condition for this to happen is that $a b c \leq a+b+c$.
Using the hypothesis $a \leq b \leq c$ we obtain the inequality
$$a b c \leq a+b+c \leq 3 c \Rightarrow a b \leq 3$$
we therefore need to consider (since $a, b$ are positive integers) the following three cases:
- $a=b=1$. Then $c$ is a divisor of $a+b=2$, and we find the first two candidate solution triples: $(a, b, c)=(1,1,1)$ and $(a, b, c)=(1,1,2)$. Both satisfy all the conditions imposed by the problem, and are thus indeed solutions.
- $a=1, b=2$. Then $c$ is a divisor of $a+b=3$ greater than or equal to $b=2$, so necessarily $c=3$ and we find the last candidate solution triple $(a, b, c)=(1,2,3)$. Indeed 1 is a divisor of $2+3=5$, 2 is a divisor of $1+3=4$ and 3 is a divisor of $1+2=3$, so the triple is a solution.
- $a=1, b=3$. Then $c$ is a divisor of $a+b=4$ greater than or equal to $b=3$, so necessarily $c=4$; but we would also need $b$ to be a divisor of $a+c$, that is, 3 a divisor of $4+1=5$, which is false. Hence the triple $(1,3,4)$ is not a solution.

By hypothesis $a+b$ is a multiple of $c$, and is less than or equal to $2 c$ (since $a \leq c, b \leq c$). We thus distinguish the cases $a+b=2 c$ and $a+b=c$.
- In the first case $2 c=a+b \leq c+c=2 c$, so in order to have equality we must have $a=b=c$. By hypothesis $a, b, c$ have greatest common divisor 1, so the only possibility is $a=b=c=1$.
- In the second case, substituting $b$ with $c-a$ in the hypothesis we obtain that $a$ divides $b+c=2 c-a$ and that $b=c-a$ divides $a+c$.
The first divisibility implies that $2 c=k a$ for some integer $k$, and since $c \geq a$ we know that $k \geq 2$. Moreover, since $c-a$ divides $c+a$, it also divides $(c+a)+(c-a)=2 c=k a$. This means that the quantity
$$\frac{k a}{c-a}=\frac{2 k a}{2 c-2 a}=\frac{2 k a}{k a-2 a}=\frac{2 k}{k-2}=\frac{2 k-4+4}{k-2}=2+\frac{4}{k-2}$$
is an integer, so $k-2$ divides 4. We know that $k \geq 2$, so $k-2$ is a non-negative divisor of 4, and hence is necessarily one of $1,2,4$.
These possibilities correspond to $k=3,4,6$, that is, to $c=\frac{3 a}{2}$, $c=2 a$, $c=3 a$ and $b=c-a=\frac{a}{2}$, $b=a$, $b=2 a$.
The first possibility is excluded by the hypothesis that $b$ be greater than or equal to $a$, while in the other two cases we find the solution triples $(a, b, c)=(a, a, 2 a)$ and $(a, 2 a, 3 a)$.
From the hypothesis that the greatest common divisor of $a, b, c$ be exactly 1 it follows that we must take $a=1$, so the only solution triples of this form are $(1,1,2)$ and $(1,2,3)$.