Note that condition (iii) can be restated as follows: iii) Given any three countries, at least two of them voted for the same problem. There are $ \binom93 = 84$ possible voting patterns for the countries. Now, let's call a triple $ \{C_1, C_2, C_3\}$, where no two countries vote for the same problem an offending triple. The total number of offending triples is $ \frac {84\binom63}{3!} = 280$. The number of offending triples that contain a voting configuration $ C$ is $ \frac {\binom63}2 = 10$. Therefore, to remove all offending triples, we need to remove at least 28 possible voting configurations out of the total 84, giving a maximum of 56 countries. This configuration is obviously possible -- say that no country voted for problem #9 (satisfying condition iii). Then there are exactly $ \binom83 = 56$ different voting patterns.