3. The numbers $1,2,3, \ldots, 29,30$ were written in a row in a random order, and partial sums were calculated: the first sum $S_{1}$ equals the first number, the second sum $S_{2}$ equals the sum of the first and second numbers, $S_{3}$ equals the sum of the first, second, and third numbers, and so on. The last sum $S_{30}$ equals the sum of all the numbers. What is the maximum possible number of odd numbers among the sums $S_{1}, S_{2}, \ldots, S_{30}$?
Answer: 23.

Solution: Evaluation: adding an odd number changes the parity of the sum, there are 15 odd numbers, so the parity of the sums changes at least 14 times. Therefore, there will be at least 7 even sums, and thus no more than 23 odd sums.

Implementation: arrange the numbers as follows: 1, then all even numbers, then all remaining odd numbers (except 1). We get the sequence: $1,2,4,6, \ldots, 30,3,5, \ldots 29$. The odd sums are the first sum (equal to 1), the next 15 (since they are formed by adding an even number to 1), and 7 of the last 14 sums $\left(S_{18}, S_{20}, \ldots, S_{30}\right)$. In total, there are $1+15+7=23$ odd sums.

Comment: The implementation is proposed - 3 points, the evaluation is made - 4 points, the points are summed. For potentially useful ideas and approaches in the absence of a solution - 2-3 points. Answer without justification - 0 points.