10.3. In a row from left to right, all natural numbers from 1 to 37 are written in such an order that each number, starting from the second to the 37th, divides the sum of all numbers to its left: the second divides the first, the third divides the sum of the first and second, and so on, the last divides the sum of the first thirty-six. The number 37 is on the first place from the left, what number is on the third place?
Answer: 2.

Solution. If the first number is the prime number 37, then the second must be 1, and the third must be a divisor of the number $37+1=38$, that is, 2 or 19. However, 19 must be in the last position, since the number 37 minus an even number must divide the sum of all the other numbers and itself, that is, divide the sum of all the numbers, which is $\frac{(1+37) \cdot 37}{2}=19 \cdot 37$, and must equal 19, as 37 is already occupied. Therefore, the number 2 is in the third position.

Grading Criteria. The solution does not require providing an example of the arrangement of numbers that satisfies the condition. If a complete correct example with the number 2 in the third position is provided without proving its uniqueness: 3 points. For constructing an example of the arrangement, if everything else is correctly proven, no additional points are added.