Determine all positive integers in the form $a<b<c<d$ with the property that each of them divides the sum of the other three.
We need to find all positive integers \(a < b < c < d\) such that each of them divides the sum of the other three. Let's denote the conditions as follows:
1. \(a \mid (b + c + d)\)
2. \(b \mid (a + c + d)\)
3. \(c \mid (a + b + d)\)
4. \(d \mid (a + b + c)\)

### Case 1: \(d = a + b + c\)
1. \(a \mid (b + c + d) = b + c + (a + b + c) = a + 2b + 2c\)
2. \(b \mid (a + c + d) = a + c + (a + b + c) = 2a + b + 2c\)
3. \(c \mid (a + b + d) = a + b + (a + b + c) = 2a + 2b + c\)

#### Subcase 1.1: \(c = 2(a + b)\)
- The numbers are \(a, b, 2(a + b), a + b + 2(a + b) = 3a + 3b\)
- We need \(b \mid (a + 2(a + b) + 3a + 3b) = 6a + 5b\), so \(b \mid 6a\)
- We also need \(a \mid (b + 2(a + b) + 3a + 3b) = 6b\), so \(a \mid 6b\)

Let \(a = a_1u\) and \(b = b_1u\) with \(\gcd(a_1, b_1) = 1\). Then \(b_1 \mid 6\) and \(a_1 \mid 6\). Possible pairs \((a_1, b_1)\) are:
- \((1, 2)\), \((1, 3)\), \((1, 6)\), \((2, 3)\), \((2, 6)\), \((3, 6)\)

Thus, the solutions are:
- \((a, b, c, d) = (u, 2u, 6u, 9u)\)
- \((a, b, c, d) = (u, 3u, 8u, 12u)\)
- \((a, b, c, d) = (u, 6u, 14u, 21u)\)
- \((a, b, c, d) = (2u, 3u, 10u, 15u)\)

#### Subcase 1.2: \(c = a + b\)
- The numbers are \(a, b, a + b, 2(a + b)\)
- We need \(b \mid (a + (a + b) + 2(a + b)) = 4a + 3b\), so \(b \mid 4a\)
- We also need \(a \mid (b + (a + b) + 2(a + b)) = 4b\), so \(a \mid 4b\)

Let \(a = a_1u\) and \(b = b_1u\) with \(\gcd(a_1, b_1) = 1\). Then \(b_1 \mid 4\) and \(a_1 \mid 4\). Possible pairs \((a_1, b_1)\) are:
- \((1, 2)\), \((1, 4)\)

Thus, the solutions are:
- \((a, b, c, d) = (u, 2u, 3u, 6u)\)
- \((a, b, c, d) = (u, 4u, 5u, 10u)\)

#### Subcase 1.3: \(c = \frac{2(a + b)}{3}\)
- The numbers are \(a, 3t - a, 2t, 5t\) with \(2t < 2a < 3t\)
- We need \(a \mid (10t - a)\), so \(a \mid 10t\)

Since \(\frac{20}{3}a < 10t < 10a\), we get:
- \(10t = 7a\) implies \(t = \frac{7a}{10}\), which is not an integer
- \(10t = 8a\) implies \(t = \frac{8a}{10} = \frac{4a}{5}\), which is not an integer
- \(10t = 9a\) implies \(t = \frac{9a}{10}\), which is not an integer

Thus, no solutions exist for this subcase.

### Case 2: \(d = \frac{a + b + c}{2}\)
1. \(a \mid (b + c + \frac{a + b + c}{2}) = \frac{3a + 3b + 3c}{2}\)
2. \(b \mid (a + c + \frac{a + b + c}{2}) = \frac{3a + 3b + 3c}{2}\)
3. \(c \mid (a + b + \frac{a + b + c}{2}) = \frac{3a + 3b + 3c}{2}\)

#### Subcase 2.1: \(c = 3(a + b)\)
- The numbers are \(a, 5t - a, 3t, 4t\) with \(4t < 2a < 5t\)
- We need \(a \mid (12t - a)\), so \(a \mid 12t\)

Since \(\frac{24}{5}a < 12t < 6a\), we get:
- \(12t = 5a\) implies \(t = \frac{5a}{12}\), which is not an integer

Thus, no solutions exist for this subcase.

#### Subcase 2.2: \(c = a + b\)
- The numbers are \(a, b, a + b, a + b\), which is not valid since \(d\) must be greater than \(c\)

Thus, no solutions exist for this subcase.

#### Subcase 2.3: \(5c = 3(a + b)\)
- The numbers are \(a, 5t - a, 3t, 4t\) with \(4t < 2a < 5t\)
- We need \(a \mid (12t - a)\), so \(a \mid 12t\)

Since \(\frac{24}{5}a < 12t < 6a\), we get:
- \(12t = 5a\) implies \(t = \frac{5a}{12}\), which is not an integer

Thus, no solutions exist for this subcase.

### Conclusion
The set of basis solutions (all others are obtained through multiplication by any positive integer) are:
- \((a, b, c, d) = (1, 2, 3, 6)\)
- \((a, b, c, d) = (1, 2, 6, 9)\)
- \((a, b, c, d) = (1, 3, 8, 12)\)
- \((a, b, c, d) = (1, 4, 5, 10)\)
- \((a, b, c, d) = (1, 6, 14, 21)\)
- \((a, b, c, d) = (2, 3, 10, 15)\)

The final answer is \( \boxed{ (1, 2, 3, 6), (1, 2, 6, 9), (1, 3, 8, 12), (1, 4, 5, 10), (1, 6, 14, 21), (2, 3, 10, 15) } \)