If there are three consecutive positive numbers $a_i$, $a_{i+1}$ and $a_{i+2}$, then $a_{i-1} > 0$. Hence we conclude that all the numbers are positive. But for the smallest number on the circle it is impossible to be greater than the sum of the next three numbers. Therefore for any three consecutive numbers, at least one of them is negative. Hence there are at least $51/3 = 17$ negative numbers on the circle. Suppose that there are exactly 17 negative numbers. Then for any three consecutive numbers only one of them is negative. Suppose that $a_i$, $a_{i+3}$, $a_{i+6}$, $a_{i+9}$, ..., where $(i = 1 \vee 2 \vee 3)$, are negative. Observe that
$$a_i > a_{i+1} + a_{i+2} + a_{i+3} > a_{i+3}.$$

Hence $a_i > a_{i+3} > a_{i+6} > \cdots > a_i$, which gives a contradiction. Therefore there are at most $51 - 18 = 33$ positive numbers on the circle. Let us give an example below.