Problem 1091

[Solution] We prove that if and only if:
(1) $k \equiv 3(\bmod 4)$, or
(2) $k \equiv 0 \quad(\bmod 4)$ and $k \geqslant 92$, $X$ can be divided into subsets $A$ and $B$ that meet the requirements.
First, we determine the conditions that $k$ should satisfy.
The sum of the elements in $X$
$$1990(k+1)+\frac{k(k+1)}{2}$$

should be even, so $4 \mid k(k+1)$.
$$k \equiv 0 \quad(\bmod 4),$$

or
$$k \equiv 3 \quad(\bmod 4).$$

If $k \equiv 0 \quad(\bmod 4)$, then $|X|=k+1$ is odd. Let $k=2m$, where $m$ is even, and
$$|A| \geqslant m+1,|B| \leqslant m,$$

At this time, the sum of the elements in $A$
$$\begin{aligned} a & \geqslant 1990+(1990+1)+\cdots+(1990+m) \\ & =1990(m+1)+\frac{m(m+1)}{2}, \end{aligned}$$
the sum of the elements in $B$
$$\begin{aligned} b & \leqslant(1990+m+1)+\cdots+(1990+2 m) \\ & =1990 m+m^{2}+\frac{m(m+1)}{2}. \end{aligned}$$

Therefore, $\quad m^{2} \geqslant 1990$.
Since $m$ is even, we have
$$m \geqslant 46, k \geqslant 92.$$

Next, we prove that the conditions are also sufficient.
Case 1: $k \equiv 3 \quad(\bmod 4)$.
In this case, the number of elements in $X$ is a multiple of 4. Since any set of 4 consecutive integers can be divided into two disjoint subsets with the same sum, $X$ has the same property.
Case 2: $k \equiv 0 \quad(\bmod 4), k \geqslant 92$.
When $k=92$, let
$$\begin{array}{l} A_{1}=\{1990,1991,1992, \cdots, 1990+46\}, \\ B_{1}=\{1990+47,1990+48, \cdots, 1990+92\}, \end{array}$$

then the sum of the elements in the latter is more by $(47-1) \times 46-1990=126$. By swapping 1990 in $A_{1}$ with 1990+63 in $B_{1}$, the resulting two sets meet the requirements (i.e., the sums of the elements are equal).
When $k>92$, by Case 1,
$$\{1990+93,1990+94, \cdots, 1990+k\}$$
(its number of elements is a multiple of 4) can be divided into two disjoint subsets with the same sum. Therefore, combining the discussion for $k=92$, $X$ has the same property.