Let $E$ be the amount of money that Eve has, and $M$ be the amount of money that Martti has. We have the following system of equations: $$\begin{cases}M+3=n(E-3)\\ E+n=3(M-n)\end{cases}$$Solving for $E$ and $M$ in terms of $n$ gives us: $$(E,M)=\left(\frac{13n+9}{3n-1},\frac{4n^2+3n+3}{3n-1}\right)$$Clearly, if $E$ is an integer, then it follows that $M$ is an integer as well. So, we only have to make sure that $E$ is an integer. We have: $$\frac{13n+9}{3n-1}\in\mathbb Z\implies\frac{39n+27}{3n-1}\in\mathbb Z\implies\frac{40}{3n-1}\in\mathbb Z$$This shows that $3n-1$ must divide $40$. The positive divisors of $40$ which are $2\pmod3$ are $2,5,8,20$. This gives us the possible values for $n$: $1,2,3,7$. Each of these gives us: $$(E,M,n)=(11,5,1),(7,5,2),(6,6,3),(5,11,7)$$Therefore, the solutions are $n=\boxed{1,2,3,7}$.