Assume $A=\{a_{1},a_{2},...,a_{12}\}$ is a set of positive integers such that for each positive integer $n \leq 2500$  there is a subset $S$ of $A$ whose sum of elements is $n$. If $a_{1}<a_{2}<...<a_{12}$ , what is the smallest possible value of $a_{1}$?
1. Given that \( A = \{a_1, a_2, \ldots, a_{12}\} \) is a set of positive integers, and for each positive integer \( n \leq 2500 \), there exists a subset \( S \) of \( A \) such that the sum of the elements of \( S \) is \( n \).
2. We need to find the smallest possible value of \( a_1 \) given that \( a_1 < a_2 < \ldots < a_{12} \).

To solve this, we need to ensure that every integer from 1 to 2500 can be represented as the sum of some subset of \( A \). 

3. Let's start by considering the smallest possible value for \( a_1 \). If \( a_1 = 1 \), then we can form the number 1 using the subset \( \{a_1\} \).

4. Next, we need to ensure that we can form every integer up to 2500. To do this efficiently, we can use the concept of binary representation. If we choose the elements of \( A \) such that each element is a power of 2, we can represent any integer up to \( 2^{12} - 1 = 4095 \) (since \( 2^{12} = 4096 \)).

5. Therefore, we can choose \( A \) as follows:
   \[
   A = \{1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048\}
   \]
   This set ensures that we can form any integer from 1 to 4095 by selecting appropriate subsets.

6. Since we need to form integers only up to 2500, the above set \( A \) is sufficient. The smallest element in this set is \( a_1 = 1 \).

Conclusion:
The smallest possible value of \( a_1 \) is \( 1 \).

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