Find all sequences of integer $x_1,x_2,..,x_n,...$ such that $ij$ divides $x_i+x_j$ for any distinct positive integer $i$, $j$.
1. **Claim**: \( x_a = x_b \) for any positive integers \( a, b \).

   **Proof**: Suppose, for the sake of contradiction, that \( x_a \neq x_b \). Consider a prime \( p \) such that \( p > |x_a - x_b| \). 

2. **Divisibility Condition**: Given that \( ij \) divides \( x_i + x_j \) for any distinct positive integers \( i, j \), we can write:
   \[
   p \mid x_p + x_a \quad \text{and} \quad p \mid x_p + x_b
   \]

3. **Subtraction**: Subtract the two congruences:
   \[
   p \mid (x_p + x_a) - (x_p + x_b) \implies p \mid x_a - x_b
   \]

4. **Contradiction**: Since \( p \) is a prime greater than \( |x_a - x_b| \), it cannot divide \( x_a - x_b \). This is a contradiction, hence our assumption that \( x_a \neq x_b \) must be false. Therefore, \( x_a = x_b \) for any positive integers \( a, b \).

5. **Constant Sequence**: Since \( x_a = x_b \) for any \( a, b \), the sequence \( \{x_i\} \) must be constant. Let \( x_i = c \) for all \( i \).

6. **Divisibility Check**: For the sequence to satisfy the given condition, \( ij \) must divide \( x_i + x_j \). Since \( x_i = x_j = c \), we have:
   \[
   ij \mid c + c \implies ij \mid 2c
   \]

7. **Conclusion**: For \( ij \) to divide \( 2c \) for all distinct \( i, j \), \( c \) must be zero. This is because \( ij \) can be arbitrarily large, and the only integer \( c \) such that \( ij \mid 2c \) for all \( i, j \) is \( c = 0 \).

Therefore, the only sequence that satisfies the given condition is the constant sequence where \( x_i = 0 \) for all \( i \).

The final answer is \( \boxed{ x_i = 0 } \) for all \( i \).