11. (Mexico 2) Determine all positive integers $k$ such that the set
$$
X=\{1990,1990+1, \cdots, 1990+k\}
$$

can be partitioned into two disjoint subsets $A$ and $B$ such that the sum of the elements in $A$ is equal to the sum of the elements in $B$.
Let the positive integer $k$ be such that
$$
X=\{1990,1990+1, \cdots, 1990+k\}
$$

can be divided into subsets $A$ and $B$ that satisfy the problem's requirements, then the sum of all elements in $X$ must be even, i.e.,
$$
\begin{array}{l}
\text { I, }(1990+n) \\
=1990(k+1), \quad h(k+1) \\
2
\end{array}
$$
This implies that $k \equiv 0 \pmod{4}$ or $k \equiv 3 \pmod{4}$.

If $k \equiv 3 \pmod{4}$, then $4 \mid (k+1)$, so we can divide $X$ such that the smallest and largest of every four consecutive integers starting from 1990 are assigned to $A$, and the other two middle ones are assigned to $B$. This way, the resulting $A$ and $B$ clearly satisfy the problem's requirements.

If $k \equiv 0 \pmod{4}$, let $k=4m$. Let $|X|, |A|, |B|$ denote the number of elements in $X, A, B$ respectively. Since $|X|=4m+1$ is odd, $|A| \neq |B|$. Without loss of generality, assume $|A| > |B|$, then $|A| \geq 2m + 1$ and $|B| \leq 2m$. This means the sum of elements in $A$ is not less than $\sum_{n=0}^{2m}(1990+n)$, and the sum of elements in $B$ is not greater than $\sum_{n=2m+1}^{4m}(1990+n)$. Thus, we have
$$
\sum_{n=0}^{2m}(1990+n) \leq \sum_{n=2m+1}^{4m}(1990+n).
$$

Therefore, $2m^2 \geq 995$, so $m \geq 23$, and $k \geq 92$.
It can be proven that when $k \equiv 0 \pmod{4}$ and $k \geq 92$, $X$ has subsets $A$ and $B$ that satisfy the problem's requirements. In fact, when $k=92$, we can let
$$
\begin{aligned}
A_1 = & \{1990, 1990+1, \cdots, 1990+46\}, \\
B_1 = & \{1990+47, 1990+49, \cdots, 1990 \\
& +92\}.
\end{aligned}
$$
However, $A_1$ and $B_1$ do not meet the problem's requirements. However, the resulting sets $A$ and $B$ do satisfy the problem's requirements.
When $k > 92$, we still assign
$$
1991, 1992, \cdots, 2036, 2053
$$

to $A$, and
1990, 2037, \cdots, 2052, 2054, \cdots, 2082

to $B$. Then, for the remaining $k-92$ numbers starting from 2083, we assign them to $A$ and $B$ alternately. Since $k \equiv 0 \pmod{4}$, $4 \mid (k-92)$, so this method will assign all $k-92$ numbers.

In summary, the positive integer $k$ satisfies the problem's requirements if and only if $k \equiv 3 \pmod{4}$, or $k \equiv 0 \pmod{4}$ and $k \geq 92$.