For a point $O$ inside a triangle $ABC$, denote by $A_1,B_1, C_1,$ the respective intersection points of $AO, BO, CO$ with the corresponding sides. Let 
\[n_1 =\frac{AO}{A_1O}, n_2 = \frac{BO}{B_1O}, n_3 = \frac{CO}{C_1O}.\] 
What possible values of $n_1, n_2, n_3$ can all be positive integers?
1. Let \( a, b, c \) be the areas of triangles \( BOC, AOC, \) and \( AOB \) respectively. The area of the entire triangle \( ABC \) is \( a + b + c \).

2. By the properties of similar triangles, we have:
   \[
   \frac{AO}{A_1O} = \frac{[BOC] + [AOC]}{[AOB]} = \frac{b + c}{a}
   \]
   Similarly,
   \[
   \frac{BO}{B_1O} = \frac{a + c}{b} \quad \text{and} \quad \frac{CO}{C_1O} = \frac{a + b}{c}
   \]

3. Define \( n_1 = \frac{AO}{A_1O} \), \( n_2 = \frac{BO}{B_1O} \), and \( n_3 = \frac{CO}{C_1O} \). Then:
   \[
   n_1 = \frac{b + c}{a}, \quad n_2 = \frac{a + c}{b}, \quad n_3 = \frac{a + b}{c}
   \]

4. Let \( x = n_1 + 1 \), \( y = n_2 + 1 \), and \( z = n_3 + 1 \). Then:
   \[
   x = \frac{a + b + c}{a}, \quad y = \frac{a + b + c}{b}, \quad z = \frac{a + b + c}{c}
   \]

5. Since \( x, y, z \) are integers, we have:
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1
   \]

6. The only integer solutions to the equation \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 \) are:
   \[
   (x, y, z) = (2, 3, 6), (3, 3, 3), (2, 4, 4)
   \]

7. Therefore, the corresponding values of \( (n_1, n_2, n_3) \) are:
   \[
   (n_1, n_2, n_3) = (1, 2, 5), (2, 2, 2), (1, 3, 3)
   \]

8. These values can be cyclically permuted, so the possible values of \( (n_1, n_2, n_3) \) are:
   \[
   (1, 2, 5), (2, 2, 2), (1, 3, 3), (2, 5, 1), (3, 3, 1), (5, 1, 2)
   \]

The final answer is \( \boxed{ (1, 2, 5), (2, 2, 2), (1, 3, 3), (2, 5, 1), (3, 3, 1), (5, 1, 2) } \).