Mathematica competitions in Croatia

The largest number of points for which Boris can be certain to gain is $503$.

We claim that if Boris arranges his cards clockwise, but in the reverse order, from $2013$ to $1$, then regardless of Andrija's arrangement of cards, Boris will gain $503$ points.



For such Boris' arrangement, let us consider some Andrija's arrangement. Under Boris' card $2013$ is Andrija's card $2k$, for some $k\in \mathbb{N}$. Let us assume that $2k\ne 2014$, the case in which $2k=2014$ follows directly. In the figure we can see the arrangement of the cards in a series, starting with the position where Boris put the card $2013$. We split pairs of cards into two parts: the part containing Andrija's cards $2k$, $2k+2$, ..., $2012$, $2014$, and the other part containing $2$, $4$, ..., $2k-4$, $2k-2$. In the first part Andrija's cards (from left to right) increase from $2k$ to $2014$, and Boris' decrease from $2013$. In other words, at the beginning of that part Boris' cards are larger, but at the end Andrija's cards are larger. Let us denote by $2a+1$ ($a\in \mathbb{N}$) the last Boris' card in the first part which is larger than the corresponding Andrija's card, which we denote by $2b$ for some $b\in \mathbb{N}$. Hence, we have $$2a+1>2b,2a-1<2b+2.$$ From the previous inequalities we get $2a-3<2b<2a+1$, so $2b=2a-2$ or $2b=2a$. Let us notice that the sum of the cards in each pair of the first part is the same and equals to $2013+2k$. In fact, in each next pair Boris' number is by $2$ smaller than the previous, while Andrija's is by $2$ larger than the previous, so the sum does not change. Hence we have $(2a+1)+2b=2013+2k$.

If $2b=2a-2$, then $k$ is odd and $a=\frac{1007+k}{2}$. Similarly, if $2b=2a$, then $k$ is even and $a=\frac{1006+k}{2}$. If $X$ is the number of the pairs in the first part in which Boris' card is larger than Andrija's, then $X=b-k+1$. Thus we have $$X=\{\begin{array}{ll}\frac{1007-k}{2},& \text{for odd }k,\\ \frac{1008-k}{2},& \text{for even }k.\end{array}$$ Furthermore, let us denote by $Y$ the number of pairs in the second part in which Boris' card is larger than Andrija's. Analogously we get $$Y=\{\begin{array}{ll}\frac{k-1}{2},& \text{for odd }k,\\ \frac{k-2}{2},& \text{for even }k.\end{array}$$ Hence $X+Y=503$, regardless of the number $k$, so we can conclude that arranging the cards in this way Boris will gain $503$ points (and Andrija $504$ points).

It remains to prove that Boris can not arrange his cards such that he gains more than $503$ points, regardless of Andrija's arrangement.

Let us assume that there is Boris' arrangement for which Andrija gains less than $504$ points, regardless of how he arranges his cards. For such Boris' arrangement, let us denote by $A$ the total number of points that Andrija gains in all of his possible arrangements. There are $1007$ possible Andrija's arrangements, so by the assumption we have $A<1007\cdot 504$. On the other hand, let us see how many points each of Boris' cards contributes to the number $A$. The card with the number $2$ contributes $1$ point because it gives points only in the arrangement in which the card with the number $1$ is on top of it. The card with the number $4$ contributes $2$ points (cases in which cards $1$ and $3$ are on top of it). In general, for $k\in \{1,2,\dots ,1007\}$ the card with the number $2k$ contributes $k$ points. Hence we have $$A=1+2+\cdots +1007=\frac{1007\cdot 1008}{2}=1007\cdot 504.$$ We reached a contradiction, so we conclude that our assumption was wrong, i.e. Boris can not be certain to gain more than $503$ points.