Solution:
Clearly for $n=1$, each pair $(x, y)$ with $x=y$ is a solution. Now, suppose that $n>1$ which implies $x \neq y$. We have
$$0 < n-1 = \frac{y^{2}+x y+3 x}{x^{2}+x y+3 y}-1 = \frac{(x+y-3)(y-x)}{x^{2}+x y+3 y}.$$
Since $x+y \geq 3$, we conclude that $x+y>3$ and $y>x$. Take $d=\operatorname{gcd}(x+y-3,\ x^{2}+x y+3 y)$. Then $d$ divides $x^{2}+x y+3 y-x(x+y-3)=3(x+y)$. Then $d$ also divides $3(x+y)-3(x+y-3)=9$, hence $d \in\{1,3,9\}$. As
$$n-1=\frac{\frac{x+y-3}{d}(y-x)}{\frac{x^{2}+x y+3 y}{d}}$$
and $\operatorname{gcd}\left(\frac{x+y-3}{d},\ \frac{x^{2}+x y+3 y}{d}\right)=1$, it follows that $\frac{x^{2}+x y+3 y}{d}$ divides $y-x$, which leads to
$$x^{2}+x y+3 y \leq d y-d x \Leftrightarrow x^{2}+d x \leq (d-3-x) y.$$
It is necessary that $d-3-x>0 \Rightarrow d>3$, therefore $d=9$ and $x<6$. Take $x+y-3=9 k,\ k \in \mathbb{N}^{*}$ since $d \mid x+y-3$ and we get $y=9 k+3-x$. Hence
$$n-1=\frac{k(9 k+3-2 x)}{k(x+3)+1}.$$
Because $k$ and $k(x+3)+1$ are relatively prime, the number $t=\frac{9 k+3-2 x}{k(x+3)+1}$ must be integer for some positive integers $x<6$. It remains to consider these values of $x$:

1) For $x=1$, then $t=\frac{9 k+1}{4 k+1}$ and since $1<t<3$, we get $t=2,\ k=1,\ y=11$, so $n=3$.

2) For $x=2$, then $t=\frac{9 k-1}{5 k+1}$ and since $1<t<2$, there are no solutions in this case.

3) For $x=3$, then $t=\frac{9 k-3}{6 k+1}$ and since $1 \neq t<2$, there are no solutions in this case.

4) For $x=4$, then $t=\frac{9 k-5}{7 k+1}<2$, i.e. $t=1$ which leads to $k=3,\ y=26$, so $n=4$.

5) For $x=5$, then $t=\frac{9 k-7}{8 k+1}<2$, i.e. $t=1$ which leads to $k=8,\ y=70$, so $n=9$.

Finally, the answer is $n \in\{1,3,4,9\}$.