Find the number of elements that a set $B$ can have, contained in $(1, 2, ... , n)$, according to the following property: For any elements $a$ and $b$ on $B$ ($a \ne b$), $(a-b) \not| (a+b)$.
1. **Understanding the Problem:**
   We need to find the maximum number of elements in a set \( B \) such that for any two distinct elements \( a \) and \( b \) in \( B \), the difference \( a - b \) does not divide the sum \( a + b \).

2. **Initial Observations:**
   - If \( a \) and \( b \) are two elements in \( B \), then \( a \neq b \).
   - We need to ensure that \( (a - b) \not| (a + b) \).

3. **Considering Differences:**
   - If \( a \) and \( b \) are consecutive integers, say \( a = k \) and \( b = k+1 \), then \( a - b = -1 \) and \( a + b = 2k + 1 \). Clearly, \(-1\) divides \( 2k + 1 \), so we cannot have consecutive integers in \( B \).
   - If \( a \) and \( b \) differ by 2, say \( a = k \) and \( b = k+2 \), then \( a - b = -2 \) and \( a + b = 2k + 2 \). Clearly, \(-2\) divides \( 2k + 2 \), so we cannot have integers differing by 2 in \( B \).

4. **Finding an Upper Bound:**
   - To avoid differences of 1 or 2, we can consider numbers that are congruent to the same residue modulo 3.
   - Let’s choose numbers that are \( 1 \mod 3 \). These numbers are of the form \( 3k + 1 \).

5. **Constructing the Set:**
   - The numbers \( 1, 4, 7, 10, \ldots \) are all \( 1 \mod 3 \).
   - The maximum number of such numbers in the range \( 1, 2, \ldots, n \) is given by \( \left\lceil \frac{n}{3} \right\rceil \).

6. **Verification:**
   - For any two numbers \( a \) and \( b \) in this set, \( a \equiv 1 \mod 3 \) and \( b \equiv 1 \mod 3 \).
   - Therefore, \( a - b \equiv 0 \mod 3 \) and \( a + b \equiv 2 \mod 3 \).
   - Since \( a - b \) is a multiple of 3 and \( a + b \) is not, \( (a - b) \not| (a + b) \).

Thus, the maximum number of elements in set \( B \) is \( \left\lceil \frac{n}{3} \right\rceil \).

The final answer is \( \boxed{ \left\lceil \frac{n}{3} \right\rceil } \).