Team Selection Test for IMO 2019

A good collection is a collection of differently numbered beads containing exactly one bead from each bag. Let $({\sigma }_1,{\sigma }_2,\dots ,{\sigma }_{2019})$ be a good collection, where for each $i=1,2,\dots ,2019$, ${\sigma }_i$ is a number of the bead taken from $i$-th bag. We say that two good collections $({\sigma }_1^{\prime },{\sigma }_2^{\prime },\dots ,{\sigma }_{2019}^{\prime })$ and $({\sigma }_1^{\prime \prime },{\sigma }_2^{\prime \prime },\dots ,{\sigma }_{2019}^{\prime \prime })$ are equivalent if for some integer $k$ and for each $i=1,2,\dots ,2019$ we have ${\sigma }_i^{\prime }+k={\sigma }_i^{\prime \prime }(\mathrm{mod}2019)$. Clearly, each good collection is equivalent to exactly $2018$ other good collections and the set of all $2019!$ good collections can be partitioned into $\frac{2019!}{2019}=2018!$ equivalence classes. The total weight of all $2019$ collections belonging to any fixed equivalence class is equal to $1\cdot 2019$ kg, since the sum representing the total weight contains the weight of each bead exactly once. Therefore, in each equivalence class there is at least one good collection with total weight not less than $2019$. Thus, in each equivalence class there is at least one proper collection and consequently there are at least $2018!$ distinct proper collections.

Now we give an example when the total number of proper collections is exactly $2018!$. Suppose that the weights of the beads of first bag are $$\frac{1}{2019}+\frac{2018}{2019^2},\frac{1}{2019}-\frac{1}{2019^2},\frac{1}{2019}-\frac{1}{2019^2},\dots ,\frac{1}{2019}-\frac{1}{2019^2}$$ respectively and the weights of all other beads are equal to $\frac{1}{2019}$. Now note that

(†) any proper collection should contain the first bead of the first bag, since otherwise its total weight is at most $1-\frac{1}{2019^2}<1$

(††) any proper collection should contain $2019$ beads, since otherwise its total weight is at most $$\frac{1}{2019}+\frac{2018}{2019^2}+2017\cdot \frac{1}{2019}=1-\frac{1}{2019^2}<1$$ Therefore, any proper collection contains the first bead of the first bag and $2018$ additional beads and the total number of proper collections is $2018!$