Eve and Martti have a whole number of euros. 
Martti said to Eve: ''If you give give me three euros, so I have $n$ times the money compared to you. '' Eve in turn said to Martti: ''If you give me $n$ euros then I have triple the amount of money compared to you'' . Suppose, that both claims are valid. 
What values can a positive integer $n$ get?
1. Let \( E \) be the amount of money that Eve has, and \( M \) be the amount of money that Martti has. We are given the following conditions:
   - If Eve gives Martti 3 euros, Martti will have \( n \) times the money compared to Eve.
   - If Martti gives Eve \( n \) euros, Eve will have triple the amount of money compared to Martti.

2. We can translate these conditions into the following system of equations:
   \[
   \begin{cases}
   M + 3 = n(E - 3) \\
   E + n = 3(M - n)
   \end{cases}
   \]

3. Solving the first equation for \( M \):
   \[
   M + 3 = nE - 3n \implies M = nE - 3n - 3
   \]

4. Substituting \( M \) from the first equation into the second equation:
   \[
   E + n = 3((nE - 3n - 3) - n)
   \]
   Simplifying inside the parentheses:
   \[
   E + n = 3(nE - 3n - 3 - n) \implies E + n = 3(nE - 4n - 3)
   \]

5. Distributing the 3 on the right-hand side:
   \[
   E + n = 3nE - 12n - 9
   \]

6. Rearranging terms to isolate \( E \):
   \[
   E + n = 3nE - 12n - 9 \implies E - 3nE = -12n - 9 - n \implies E(1 - 3n) = -13n - 9
   \]

7. Solving for \( E \):
   \[
   E = \frac{-13n - 9}{1 - 3n}
   \]

8. Substituting \( E \) back into the expression for \( M \):
   \[
   M = nE - 3n - 3 = n \left( \frac{-13n - 9}{1 - 3n} \right) - 3n - 3
   \]
   Simplifying the expression:
   \[
   M = \frac{n(-13n - 9)}{1 - 3n} - 3n - 3 = \frac{-13n^2 - 9n}{1 - 3n} - 3n - 3
   \]
   Combining the terms:
   \[
   M = \frac{-13n^2 - 9n - 3n(1 - 3n) - 3(1 - 3n)}{1 - 3n} = \frac{-13n^2 - 9n - 3n + 9n^2 - 3 + 9n}{1 - 3n} = \frac{-4n^2 - 3}{1 - 3n}
   \]

9. For \( E \) to be an integer, the denominator \( 1 - 3n \) must divide the numerator \( -13n - 9 \). This implies:
   \[
   \frac{13n + 9}{3n - 1} \in \mathbb{Z} \implies \frac{39n + 27}{3n - 1} \in \mathbb{Z} \implies \frac{40}{3n - 1} \in \mathbb{Z}
   \]

10. Therefore, \( 3n - 1 \) must be a divisor of 40. The positive divisors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40. We need to check which of these satisfy \( 3n - 1 \equiv 0 \pmod{3} \):
    - \( 3n - 1 = 1 \implies n = \frac{2}{3} \) (not an integer)
    - \( 3n - 1 = 2 \implies n = 1 \)
    - \( 3n - 1 = 4 \implies n = \frac{5}{3} \) (not an integer)
    - \( 3n - 1 = 5 \implies n = 2 \)
    - \( 3n - 1 = 8 \implies n = 3 \)
    - \( 3n - 1 = 10 \implies n = \frac{11}{3} \) (not an integer)
    - \( 3n - 1 = 20 \implies n = 7 \)
    - \( 3n - 1 = 40 \implies n = 14 \) (not a valid solution as it does not satisfy the conditions)

11. Therefore, the possible values for \( n \) are \( 1, 2, 3, 7 \).

The final answer is \( n = \boxed{1, 2, 3, 7} \)