Find all positive integers $n$ such that the equation $y^{2}+x y+3 x=n\left(x^{2}+x y+3 y\right)$ has at least a solution $(x, y)$ in positive integers.
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 $03$ 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 $n \in\{1,3,4,9\}$.