Let the sum of positive numbers $x_1, x_2, ... , x_n$ be $1$. 
Let $s$ be the greatest of the numbers $\big\{\frac{x_1}{1+x_1}, \frac{x_2}{1+x_1+x_2}, ..., \frac{x_n}{1+x_1+...+x_n}\big\}$. 
What is the minimal possible $s$? What $x_i $correspond it?
1. We are given that the sum of positive numbers \( x_1, x_2, \ldots, x_n \) is 1, i.e., 
   \[
   x_1 + x_2 + \cdots + x_n = 1.
   \]
2. We need to find the minimal possible value of \( s \), where \( s \) is the greatest of the numbers 
   \[
   \left\{ \frac{x_1}{1+x_1}, \frac{x_2}{1+x_1+x_2}, \ldots, \frac{x_n}{1+x_1+\cdots+x_n} \right\}.
   \]
3. To find the minimal possible \( s \), we need to analyze the expression \( \frac{x_i}{1 + \sum_{j=1}^i x_j} \).

4. Consider the function \( f(x) = \frac{x}{1+x} \). This function is increasing for \( x > 0 \) because its derivative \( f'(x) = \frac{1}{(1+x)^2} \) is positive for all \( x > 0 \).

5. Since \( f(x) \) is increasing, the maximum value of \( \frac{x_i}{1 + \sum_{j=1}^i x_j} \) will occur for the largest \( x_i \) and the smallest denominator \( 1 + \sum_{j=1}^i x_j \).

6. To minimize \( s \), we need to distribute the \( x_i \) values such that the maximum value of \( \frac{x_i}{1 + \sum_{j=1}^i x_j} \) is minimized.

7. Suppose \( x_1 = x_2 = \cdots = x_n = \frac{1}{n} \). Then,
   \[
   \sum_{j=1}^i x_j = \frac{i}{n}.
   \]
   Therefore,
   \[
   \frac{x_i}{1 + \sum_{j=1}^i x_j} = \frac{\frac{1}{n}}{1 + \frac{i}{n}} = \frac{1}{n + i}.
   \]

8. The greatest value of \( \frac{1}{n + i} \) occurs when \( i \) is smallest, i.e., \( i = 1 \). Thus,
   \[
   s = \frac{1}{n + 1}.
   \]

9. To verify that this is indeed the minimal possible \( s \), consider any other distribution of \( x_i \). If \( x_i \) are not equal, then some \( x_i \) will be larger than \( \frac{1}{n} \), and the corresponding \( \frac{x_i}{1 + \sum_{j=1}^i x_j} \) will be larger than \( \frac{1}{n + 1} \).

10. Therefore, the minimal possible \( s \) is achieved when \( x_1 = x_2 = \cdots = x_n = \frac{1}{n} \).

The final answer is \( \boxed{\frac{1}{n+1}} \).