4. On equal-armed scales, weights of $1, 2, 3, \ldots, n, n>6$ are placed in such a way that the scales are in equilibrium. Is it always possible to remove 3 weights from the scales in such a way that the scales remain in equilibrium?
Solution. If for some $k>1$ one weight $k+1$ is placed on the first scale pan, and on the second - weights $k$ and 1, then it is sufficient to remove these 3 weights and the requirement is met. If such a situation does not occur, then consider the scale pan where the weight 1 is located, as well as the weight with the smallest weight greater than 1. Let this weight be $k$. Then on the second pan, there must be all weights with masses $2,3, \ldots, k-1$. On the first, there must be the weight $k+1$. If it were on the second pan, then it would be possible to remove from the first pan the weights 1 and $k$, and from the second the weight $k+1$ and the scales would remain in balance (the first case). Note that, $kn+1$. Therefore, the weights are distributed on the pans as follows: On the first $-1, k, k+1, \ldots$, on the second $-2,3, \ldots, k-1, \ldots$. From this, it is clear that $k>3$, otherwise, the equality would not hold. But then on one pan there are weights $2, k-1$, and on the other - the weight $k+1$, which can be removed.

Answer: Always.