We need to find all pairs $(a, b)$ of positive integers such that:
1. $2a-1$ and $2b+1$ are coprime,
2. $a+b$ divides $4ab+1$.

### Step 1: Analyze the Conditions

Condition 1:
The integers $2a-1$ and $2b+1$ are coprime, meaning their greatest common divisor (GCD) is 1. Therefore:

$$\gcd(2a-1, 2b+1) = 1$$

This condition ensures that $2a-1$ and $2b+1$ have no common factors other than 1.

Condition 2:
The sum $a+b$ divides $4ab+1$, expressed as:

$$a+b \mid 4ab+1$$

This implies that there exists an integer $k$ such that:

$$4ab + 1 = k(a+b)$$

Rearranging gives:

$$4ab + 1 = ka + kb \quad \Rightarrow \quad 4ab + 1 - ka - kb = 0 \quad \Rightarrow \quad (4a - k)b = ka - 1$$

From this expression, find conditions that will satisfy both divisibility and coprimacy.

### Step 2: Find Possible Solutions

Let's assume $b = a + 1$, and we will verify whether this satisfies both conditions:

1. Coprime Check:
- $2a-1$ becomes $2a-1$.
- $2b+1$ becomes $2(a+1)+1 = 2a + 3$.

We check for coprimacy:

$$\gcd(2a-1, 2a+3) = \gcd(2a-1, 4) = 1$$

Thus, they are coprime when the GCD is 1.

2. Divisibility Check:
- $a + b = a + (a + 1) = 2a + 1$
- This divisibility condition becomes:

$$2a + 1 \mid 4ab + 1$$

Substituting $b = a + 1$ gives:

$$4a(a+1) + 1 = 4a^2 + 4a + 1$$

Check:

$$4a^2 + 4a + 1 \div (2a + 1) = 2a + 1$$

The quotient is an integer, so this condition satisfies the divisibility requirement under the assumption.

Therefore, the pair $(a, a+1)$ satisfies all conditions.

### Conclusion

From the steps above, the only pairs $(a, b)$ that satisfy the given conditions are:

$$\boxed{(a, a+1)}$$

This solution set meets both the coprimacy and divisibility requirements.