Answer: 198. Solution: Let the consecutive numbers be $a, b, c, d, e$. Notice that if $a > c$, then $a + b > b + c$, which implies $c < a$. Similarly, if $c d > d e$, then $c > e$. Continuing this reasoning, we get that each number is greater than the number that is two positions clockwise from it. However, writing a chain of 99 such inequalities starting from some number $a$, we get that $a > a$ - a contradiction. Therefore, all 99 placed numbers must be equal to the same number $a$, such that $2a = a^2$, from which $a = 2$, and the desired sum is $2 \times 99 = 198$.
5. Ninety-nine positive numbers are arranged in a circle. It turns out that for any four consecutive numbers, the sum of the first two of them in a clockwise direction is equal to the product of the last two of them in a clockwise direction. What can the sum of all 99 arranged numbers be? (S. Berlov)