Problem 1056

[Solution] We prove that it is impossible to divide the set of integers into 3 subsets such that $n, n-50, n+1987$ belong to 3 different subsets, where $n$ is any integer.

If a 3-tuple of 3 numbers each comes from 3 different subsets, then we call this 3-tuple "representative."

Assume there exists a partition that meets the requirements of the problem, then for any integer $n$, the following 3-tuples are "representative":
$$\begin{array}{l} (n-50, \quad n, \quad n+1987), \\ (n-100, \quad n-50, \quad n+1937) \\ (n+1937, \quad n+1987, \quad n+2 \cdot 1987) . \end{array}$$

We use the notation $i \sim j$ to indicate that the integer $i$ and the integer $j$ belong to the same subset, and the notation $i \propto j$ to indicate that the integer $i$ and the integer $j$ do not belong to the same subset.
From (2) we know
$$n+1937 \propto n-50 \text {, }$$

From (3) we know
$$n+1937 \propto n+1987$$

From (1) we know
$$\begin{array}{r} n \propto n-50, \\ n \propto n+1987 . \end{array}$$

Thus, it can only be that
$$n \sim n+1937 \text {. }$$

From (2), by replacing $n+1937$ with $n$, we get another "representative" 3-tuple
$$(n-100, n-50, n) \text {. }$$

We can also get (by replacing $n$ with $n-50$)
$$(n-150, \quad n-100, \quad n-50) \text {. }$$

which is also a "representative" tuple.
Thus, from (5) and (6) we know
$$n \propto n-50, \quad n \propto n-100 \text {, }$$

Thus, it can only be that
$$n \sim n-150 \text {. }$$

From (4) and (7) we get
$$\begin{array}{c} 0 \sim 1937 \sim 2 \cdot 1937 \sim \cdots \sim 50 \cdot 1937 \\ =646 \cdot 150-50 \sim 645 \cdot 150-50 \sim \cdots \sim-50 . \end{array}$$

Thus $0 \sim-50$, i.e., when $n=0$, $n$ and $n-50$ belong to the same subset, leading to a contradiction. Therefore, there cannot be a partition that meets the requirements of the problem.