Find the maximum number of elements which can be chosen from the set $ \{1,2,3,\ldots,2003\}$ such that the sum of any two chosen elements is not divisible by 3.
1. **Classify the elements of the set $\{1, 2, 3, \ldots, 2003\}$ based on their residues modulo 3.**
   - Any integer $n$ can be written in one of the forms: $3k$, $3k+1$, or $3k+2$ for some integer $k$.
   - Therefore, the elements of the set can be classified into three categories:
     - Numbers congruent to $0 \pmod{3}$: $\{3, 6, 9, \ldots, 2001\}$
     - Numbers congruent to $1 \pmod{3}$: $\{1, 4, 7, \ldots, 2002\}$
     - Numbers congruent to $2 \pmod{3}$: $\{2, 5, 8, \ldots, 2003\}$

2. **Count the number of elements in each category.**
   - Numbers congruent to $0 \pmod{3}$:
     \[
     3k \leq 2003 \implies k \leq \left\lfloor \frac{2003}{3} \right\rfloor = 667
     \]
     So, there are 667 elements congruent to $0 \pmod{3}$.
   - Numbers congruent to $1 \pmod{3}$:
     \[
     3k + 1 \leq 2003 \implies 3k \leq 2002 \implies k \leq \left\lfloor \frac{2002}{3} \right\rfloor = 667
     \]
     So, there are 668 elements congruent to $1 \pmod{3}$.
   - Numbers congruent to $2 \pmod{3}$:
     \[
     3k + 2 \leq 2003 \implies 3k \leq 2001 \implies k \leq \left\lfloor \frac{2001}{3} \right\rfloor = 667
     \]
     So, there are 668 elements congruent to $2 \pmod{3}$.

3. **Determine the maximum number of elements that can be chosen such that the sum of any two chosen elements is not divisible by 3.**
   - The sum of two numbers is divisible by 3 if:
     - Both numbers are congruent to $0 \pmod{3}$.
     - One number is congruent to $1 \pmod{3}$ and the other is congruent to $2 \pmod{3}$.
   - To avoid sums that are divisible by 3, we can choose:
     - At most one element from the set of numbers congruent to $0 \pmod{3}$.
     - All elements from either the set of numbers congruent to $1 \pmod{3}$ or the set of numbers congruent to $2 \pmod{3}$, but not both.

4. **Calculate the maximum number of elements that can be chosen.**
   - We can choose at most one element from the set of numbers congruent to $0 \pmod{3}$.
   - We can choose all 668 elements from either the set of numbers congruent to $1 \pmod{3}$ or the set of numbers congruent to $2 \pmod{3}$.
   - Therefore, the maximum number of elements that can be chosen is:
     \[
     1 + 668 = 669
     \]

The final answer is $\boxed{669}$.