Assume Adrian gets rid of the $n$ consecutive cards numbered $k,k+1, \ldots ,k+n-1$. Consequently the sum of the numbers on remaing $n$ cards is $S = \sum_{m=1}^{2n} m - \sum_{m=0}^{n-1} (k+m) = {\textstyle \frac{2n(2n + 1)}{2} - \big( kn + \frac{n(n - 1)}{2} \big)}$. i.e. $(1) \;\; 2S = n(3n + 3 - 2k)$, which give us ${\textstyle (2) \;\; k = \frac{1}{2}[3(n + 1) - \frac{2S}{n}]}$, where $(3) \;\; 1 \leq k \leq n+1$. Combining formula (2) and condition (3) with the fact that $S=1615$, we find that $n(n + 1) \leq 3230 < 3n(n + 1)$, yielding $(4) \;\; 33 \leq n \leq 56$. Moreover by formula (1) $n \mid 2S = 2 \cdot 1615 = 2 \cdot 5 \cdot 17 \cdot 19$, which combined with condition (4) give us $n = 2 \cdot 17 = 34$ or $n = 2 \cdot 19 = 38$.