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 $2k \neq 2014$, 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
$$2a+1 > 2b, \quad 2a-1 < 2b+2.$$
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
$$X = \begin{cases} \frac{1007-k}{2}, & \text{for odd } k, \\ \frac{1008-k}{2}, & \text{for even } k. \end{cases}$$
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
$$Y = \begin{cases} \frac{k-1}{2}, & \text{for odd } k, \\ \frac{k-2}{2}, & \text{for even } k. \end{cases}$$
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
$$A = 1 + 2 + \dots + 1007 = \frac{1007 \cdot 1008}{2} = 1007 \cdot 504.$$
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.