Two of the numbers $a+b, a-b, ab, a/b$ are positive, the other two are negative. Find the sign of $b$
To determine the sign of \( b \), we need to analyze the given conditions and the relationships between the expressions \( a+b, a-b, ab, \frac{a}{b} \).

1. **Analyzing the product \( ab \) and the quotient \( \frac{a}{b} \):**
   \[
   ab \cdot \frac{a}{b} = a^2
   \]
   Since \( a^2 > 0 \) for any non-zero \( a \), it follows that \( ab \) and \( \frac{a}{b} \) must have the same sign.

2. **Case 1: \( ab > 0 \)**
   - If \( ab > 0 \), then both \( a \) and \( b \) must have the same sign (both positive or both negative).

   - **Subcase 1.1: Both \( a \) and \( b \) are positive**
     - If \( a \) and \( b \) are both positive, then \( a+b > 0 \) and \( a-b \geq 0 \). However, this contradicts the condition that two of the numbers are positive and two are negative because \( a+b \) and \( a-b \) would both be non-negative.

   - **Subcase 1.2: Both \( a \) and \( b \) are negative**
     - If \( a \) and \( b \) are both negative, then \( a+b < 0 \) and \( a-b \leq 0 \). This also contradicts the condition that two of the numbers are positive and two are negative because \( a+b \) and \( a-b \) would both be non-positive.

3. **Case 2: \( ab < 0 \)**
   - If \( ab < 0 \), then \( a \) and \( b \) must have opposite signs (one positive and one negative).

   - **Subcase 2.1: \( a > 0 \) and \( b < 0 \)**
     - If \( a > 0 \) and \( b < 0 \), then:
       \[
       a+b < 0 \quad \text{(since } b \text{ is negative)}
       \]
       \[
       a-b > 0 \quad \text{(since } b \text{ is negative)}
       \]
       This satisfies the condition that two of the numbers are positive and two are negative.

   - **Subcase 2.2: \( a < 0 \) and \( b > 0 \)**
     - If \( a < 0 \) and \( b > 0 \), then:
       \[
       a+b < 0 \quad \text{(since } a \text{ is negative)}
       \]
       \[
       a-b < 0 \quad \text{(since } a \text{ is negative)}
       \]
       This does not satisfy the condition that two of the numbers are positive and two are negative because both \( a+b \) and \( a-b \) would be negative.

From the analysis, the only scenario that satisfies the given conditions is when \( a > 0 \) and \( b < 0 \).

The final answer is \( \boxed{ b < 0 } \)