National Olympiad of Argentina

The answer is 14. First we prove that no collection of 13 weights is admissible. Assume on the contrary that 13 weights $a_1,a_2,\dots ,a_{13}$ can be divided into 9, 5 and 4 groups with equal masses (divisions 1, 2 and 3 respectively). Multiplying all $a_i$ by a positive number yields an admissible collection again, so suppose that the total mass is $180=4\cdot 5\cdot 9$. (It is not assumed here that the $a_i$ are integers.) The mass of one group in division 1 is $180/9=20$, hence $a_i\le 20$ for all $i$. At least 5 groups of the 9 groups in division 1 consist of exactly 1 weight, so there are 5 weights of mass 20, say $a_1,\dots ,a_5$. They are in different groups in division 2 where the mass of one group is $180/5=36$. Remove $a_1,\dots ,a_5$ from their groups to obtain a division of $a_6,\dots ,a_{13}$ into 5 groups of mass 16. In particular $a_i\le 16$ for $6\le i\le 13$. So in division 1 the 8 weights $a_6,\dots ,a_{13}$ are divided into 4 pairs with mass 20 each, say $\{a_6,a_7\},\{a_8,a_9\},\{a_{10},a_{11}\},\{a_{12},a_{13}\}$. There is also a division of $a_6,\dots ,a_{13}$ into 5 groups of mass 16; two of these groups must have exactly one weight. Hence one may assume $a_6=a_8=16$, implying $a_7=a_9=4$. All $a_i$ determined so far are integer multiples of 4.

Each group in division 3 has mass $180/4=45$ which is not a multiple of 4. Hence each group contains a non-multiple of 4, i.e. an integer not divisible by 4 or a non-integer. As $a_1,\dots ,a_9$ are divisible by 4, the non-multiples of 4 are $a_{10},a_{11},a_{12},a_{13}$, and they belong to different groups. We obtain that each of $a_{10},a_{11},a_{12},a_{13}$ differs from 45 by a multiple of 4, so it is an integer congruent to 1 modulo 4. But then $a_{10}+a_{11}\ne 20$ which is a contradiction.

[The reasoning modulo 4 can be avoided. We proved that weights $a_6,\dots ,a_{13}$ form 5 groups of mass 16—hence each one has mass at most 16—and also 4 pairs with mass 20—so each one has mass at least 4. Two of the 5 weights with mass 20 are in the same group in division 3. The mass 45 of the group is completed by exactly one weight of mass 5: having two or more weights of total mass 5 would yield a mass less than 4. The weight 5 must be paired up with a weight 15 in a group of mass 20. However the 15 is also in a group of mass 16 which is impossible by the above.]

So there is no admissible collection with 13 weights (and hence with fewer weights either). An admissible collection with 14 weights is $3,4,5,7,9,11,13,15,16,17,20,20,20,20$; divisions 1, 2 and 3 are:

$\{3,17\},\{4,16\},\{5,15\},\{7,13\},\{9,11\},\{20\},\{20\},\{20\},\{20\}$;

$\{3,13,20\},\{4,15,17\},\{5,11,20\},\{7,9,20\},\{16,20\}$;

$\{3,4,7,11,20\},\{5,20,20\},\{9,16,20\},\{13,15,17\}$.