Problem 1543

1. Define the points and distances:
Let the four points be $A, B, C, D$ in order from left to right. Define the distances between consecutive points as follows:
$$AB = x, \quad BC = y, \quad CD = z$$

2. Calculate the sum of distances from each point:
- For point $A$:
$$\text{Sum of distances} = AB + AC + AD = x + (x + y) + (x + y + z) = 3x + 2y + z$$
- For point $B$:
$$\text{Sum of distances} = BA + BC + BD = x + y + (y + z) = x + 2y + z$$
- For point $C$:
$$\text{Sum of distances} = CA + CB + CD = (x + y) + y + z = x + 2y + z$$
- For point $D$:
$$\text{Sum of distances} = DA + DB + DC = (x + y + z) + (y + z) + z = x + 2y + 3z$$

3. Identify the necessary conditions:
- From the calculations, we see that the sums of distances from $B$ and $C$ are equal:
$$\text{Sum from } B = \text{Sum from } C = x + 2y + z$$
- The sums of distances from $A$ and $D$ are:
$$\text{Sum from } A = 3x + 2y + z$$
$$\text{Sum from } D = x + 2y + 3z$$

4. Check each option:
- **Option (a) $29, 29, 35, 37$:**
- We need two equal numbers, which are $29$ and $29$. Assign these to $B$ and $C$:
$$x + 2y + z = 29$$
- The remaining sums are $35$ and $37$. Assign these to $A$ and $D$:
$$3x + 2y + z = 35$$
$$x + 2y + 3z = 37$$
- Solve the system of equations:
$$x + 2y + z = 29 \quad \text{(1)}$$
$$3x + 2y + z = 35 \quad \text{(2)}$$
$$x + 2y + 3z = 37 \quad \text{(3)}$$
- Subtract (1) from (2):
$$3x + 2y + z - (x + 2y + z) = 35 - 29$$
$$2x = 6 \implies x = 3$$
- Subtract (1) from (3):
$$x + 2y + 3z - (x + 2y + z) = 37 - 29$$
$$2z = 8 \implies z = 4$$
- Substitute $x = 3$ and $z = 4$ back into (1):
$$3 + 2y + 4 = 29$$
$$2y = 22 \implies y = 11$$
- Therefore, $x = 3$, $y = 11$, $z = 4$ satisfies all conditions. Thus, option (a) is possible.

- **Option (b) $28, 29, 35, 37$:
- We need two equal numbers, but there are no two equal numbers in this set. Thus, option (b) is not possible.

- Option (c) $28, 34, 34, 37$:**
- We need two equal numbers, which are $34$ and $34$. Assign these to $B$ and $C$:
$$x + 2y + z = 34$$
- The remaining sums are $28$ and $37$. Assign these to $A$ and $D$:
$$3x + 2y + z = 28$$
$$x + 2y + 3z = 37$$
- Solve the system of equations:
$$x + 2y + z = 34 \quad \text{(1)}$$
$$3x + 2y + z = 28 \quad \text{(2)}$$
$$x + 2y + 3z = 37 \quad \text{(3)}$$
- Subtract (1) from (2):
$$3x + 2y + z - (x + 2y + z) = 28 - 34$$
$$2x = -6 \implies x = -3$$
- Subtract (1) from (3):
$$x + 2y + 3z - (x + 2y + z) = 37 - 34$$
$$2z = 3 \implies z = 1.5$$
- Substitute $x = -3$ and $z = 1.5$ back into (1):
$$-3 + 2y + 1.5 = 34$$
$$2y = 35.5 \implies y = 17.75$$
- Since $x = -3$, $y = 17.75$, $z = 1.5$ are not valid distances (distances cannot be negative), option (c) is not possible.

Conclusion:
Only option (a) $29, 29, 35, 37$ is possible.

The final answer is $\boxed{ 29, 29, 35, 37 }$