Removing pieces of paper

Total sum of all pages is $2n(2n+1)/2=n(2n+1)$

Let the papers Albert took is starts with the positive integer $k$ .Then consecutive $n$ papers are $k,k+1,k+2,.....,k+n-1$ .Sum of these papers is $nk+n(n-1)/2$ .

According to the question $n(2n+1) -nk-n(n-1)/2=1615$ .Re-arranging the equation

$n(3n+3 - 2 k)=2 \times 5 \times 17 \times 19$

If $n$ is odd then $3n+3$ is even and $3n+3 - even=2k$ so, $2k$ will be even and $k$ will be an integer. ..But we have to check $k+n-1 \leq 2n$ .And it is easy to check no odd value will be allowable.

If $n$ is even $3n+3$ is odd and $3n+3-odd=2k$ .Again $k$ will be an integer.Here $k$ satisfy the above condition that is $k+n-1 \leq 2n$ .So, $n=34,38$ are only solutions.