II. (50 points) Given the set $X=\{2006,2006+$ $1,2006+2, \cdots, 2006+k\}$. Find the positive integer $k$, such that the set $X$ can be partitioned into three sets $A$, $B$, and $C$, satisfying:
(1) $A \cup B \cup C=X$,
$A \cap B=B \cap C=C \cap A=\varnothing$;
(2) The sum of the elements in sets $A$, $B$, and $C$ are equal.
(ii) When $m$ is even, i.e., $k=6n-1\left(n \in \mathbf{N}_{+}\right)$, the number of elements in set $X$ is a multiple of 6. It is only necessary to start from 2006, and in every 6 consecutive odd numbers, include the first two numbers in $A$, the middle two numbers in $B$, and the remaining two numbers in $C$. Thus, the sets $A$, $B$, and $C$ obtained in this way meet the requirements.
(ii) When $m$ is odd, i.e., $k=6n-4\left(n \in \mathbf{N}_{+}\right)$, if $n=1$, $X=\{2006,2006+1,2006+2\}$, it is clear that there is no division that meets the conditions.
If $n=2$, take
$$
\begin{array}{l}
A=\{2006,2006+4,2006+8\}, \\
B=\{2006+2,2006+3,2006+7\}, \\
C=\{2006+1,2006+5,2006+6\}
\end{array}
$$

This is a division that meets the conditions.
If $n=3$, the set $X$ has 6 more elements than when $n=2$. By incorporating the additional 6 elements into the sets $A$, $B$, and $C$ as described in (i), the resulting sets meet the requirements.
By analogy, for $n>3$, a division that meets the conditions exists.
In summary, the positive integers sought are
$$
k=3m-1\left(m \in \mathbf{N}_{+}, m>1\right) .
$$