15. (MEX 2) Determine for which positive integers $k$ the set
$$
X=\{1990,1990+1,1990+2, \ldots, 1990+k\}
$$
can be partitioned into two disjoint subsets $A$ and $B$ such that the sum of the elements of $A$ is equal to the sum of the elements of $B$.
15. Let $S(Z)$ denote the sum of all the elements of a set $Z$. We have $S(X)=$ $(k+1)-1990+\frac{k(k+1)}{2}$. To partition the set into two parts with equal sums, $S(X)$ must be even and hence $\frac{k(k+1)}{2}$ must be even. Hence $k$ is of the form $4 r$ or $4 r+3$, where $r$ is an integer.
For $k=4 r+3$ we can partition $X$ into consecutive fourtuplets $\{1990+$ $4 l, 1990+4 l+1,1990+4 l+2,1990+4 l+3\}$ for $0 \leq l \leq r$ and put $1990+4 l, 1990+4 l+3 \in A$ and $1990+4 l+1,1990+4 \bar{l}+2 \in B$ for all $l$. This would give us $S(A)=S(B)=(3980+4 r+3)(r+1)$.
For $k=4 r$ the numbers of elements in $A$ and $B$ must differ. Let us assume without loss of generality $|A|<|B|$. Then $S(A) \leq(1990+2 r+1)+(1990+$ $2 r+2)+\cdots+(1990+4 r)$ and $S(B) \geq 1990+1991+\cdots+(1990+2 r)$. Plugging these inequalities into the condition $S(A)=S(B)$ gives us $r \geq 23$ and consequently $k \geq 92$. We note that $B=\{1990,1991, \ldots, 2034,2052,2082\}$ and $A=\{2035,2036, \ldots, 2051,2053, \ldots, 2081\}$ is a partition for $k=92$ that satisfies $S(A)=S(B)$. To construct a partition out of higher $k=4 r$ we use the $k=92$ partition for the first 93 elements and construct for the remaining elements as was done for $k=4 x+3$.
Hence we can construct a partition exactly for the integers $k$ of the form $k=4 r+3, r \geq 0$, and $k=4 r, r \geq 23$.