It can be shown that the losing positions are precisely those of the form $(x,x,y)$ where $x=y$ or $\nu_2(x-y)$ is even. Can anyone please tell me if there is a mistake (and where is it) in this approach by @IvayloVasilev? Suppose for contradiction that $(2000,4000,4899)$ is losing. Then $(2899, 4000, 4000)$ is winning, so $(2899+x,4000,4000-x)$ is losing for some $x$. However, by transferring $899+x$ in the initial position we get that $(2899+x, 4000, 4000-x)$ is winning, contradiction. Therefore $(2000,4000,4899)$ is winning. UPDATE: Several people confirmed there is no error - this argument just shows that $(x,y,z)$ for distinct $x,y,z$ is a winning position, it just does not give any info on $(x,x,y)$ positions, which does not matter for the aim of the problem. It is just that apriori I did not expect such a straightforward solution to exist, having in mind the original source (and official solution) of this problem...