The product of 50 consecutive integers is zero and the sum of these numbers is positive. What is the smallest value that this sum can assume?
As the product is equal to zero, one of the numbers must be zero. Thus, to minimize the sum, we should have the largest number of negative numbers, but in such a way that the sum is still positive.

Therefore, the number of negative numbers must be less than the number of positive numbers. Thus, among the 49 non-zero numbers, 24 are negative and 25 are positive. Therefore, the minimum sum is

$$
\begin{gathered}
-24+(-23)+(-22)+\cdots+(-1)+0+1+\cdots+25= \\
25+(-24+24)+(-23+23)+\cdots+(-1+1)+0=25
\end{gathered}
$$

Suggestion: Observe that from the number 777 onwards, all numbers leave the same remainder when divided by 1000.
Suggestion: Observe the number of times a person's age is considered in the ten sums.
Suggestion: If the product of the numbers is equal to zero, then one of the numbers must be equal to zero.

Suggestion: Observe what happens to the sum of the digits of the number when the operation described in the problem is performed.

Facts that Help: The arithmetic mean of two numbers \(a\) and \(b\) is given by

$$
\frac{a+b}{2}
$$

(1) \(2(3) 4 \quad 5 \quad 6 \quad 7 \quad 8 \quad 9\)

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(6) 2203401689

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\(\begin{array}{lllllllll}8 & 2 & 2 & 3 & 4 & 5 & 6 & 7 & 8\end{array}\)

Suggestion: Analyze the remainders of the numbers in the sequence when divided by 3.

Facts that Help: A number and the sum of its digits leave the same remainder when divided by 3.