OP should add : "Every evening the thief stops trusting one of his accomplices." solution: If a policeman shadows someone not trusted by the thief, he will never catch the thief; let us call such policeman redundant. We prove that the thief can use a greedy algorithm, dismissing every evening an accompilce to make more policemen redundant. On Day $k,$ the thief can dismiss any of $2 m-k+1$ accomplices, and each useful policeman can be made redundant in $k$ ways. Therefore in this step the ratio of policemen which can be made redundant is at least $\frac{k}{2 m-k+1} .$ It remains to note that $$ \prod_{k=1}^{m}\left(1-\frac{k}{2 m-k+1}\right)=\frac{2 m-1}{2 m} \cdot \frac{2 m-3}{2 m-1} \cdot \frac{2 m-5}{2 m-2} \cdot \ldots \cdot \frac{1}{m+1} $$Multiplying the numerator and the denominator by $$ m !=2^{-m} \cdot 2 m \cdot(2 m-2) \cdot \ldots \cdot 2 $$we see that the product equals $2^{-m}$. Thus if the number of policemen is less than $2^{m},$ all of them become redundant at some moment.