Solution:

We will show that there is a unique solution, namely $n=2700$.

Let $n$ be a positive integer satisfying the given conditions. We immediately observe that, since $n$ is a multiple of $300$, and hence of $100$, the units and tens digits of $n$ must be equal to zero. Suppose then that its decimal representation is $n=a_{k} a_{k-1} \cdots a_{2} 00$, where, as usual, the first digit $a_{k}$ is different from zero. We have
$$n=10^{k} a_{k}+10^{k-1} a_{k-1}+\cdots+100 a_{2} \geq 10^{k}$$
and, since all digits are less than or equal to $9$,
$$s:=a_{k}+a_{k-1}+\cdots+a_{2} \leq 9(k-1) .$$
From the hypothesis $n=300 s$ we obtain, by the preceding inequalities, $10^{k} \leq 300 \cdot 9(k-1)=2700(k-1)$. We verify by induction that this last inequality is false for $k \geq 4$: the base step is the verification that $10^{4}>2700 \cdot 3=8100$; for the inductive step, suppose that $10^{k}>2700(k-1)$: then $10^{k+1}>10^{k} \cdot 2>2700(2 k-2)>2700(k-2)$.

It follows that $k \leq 3$. Thus $n$ is of the form $10^{3} a_{3}+10^{2} a_{2}$ (in this case $a_{3}=0$ would a priori also be allowed, although as we shall see this does not occur). The equation $n=300 s$ becomes $1000 a_{3}+100 a_{2}=300 a_{3}+300 a_{2}$, which in turn simplifies to
$$7 a_{3}=2 a_{2} .$$
From this last equation we deduce that $7$ is a divisor of $a_{2}$, so the only possibility for having a positive integer is $a_{2}=7$ and consequently $a_{3}=2$. On the other hand, the number thus found, $n=2700$, does indeed satisfy the conditions of the problem.