The following works. I'll use a shorter example than you, not so much to write, we will use \[ a = \begin{bmatrix} 1 & 2 \\ 3 & 6 \\ 9 & 13 \end{bmatrix} \] First we will add a first column contating the old first column -1 and a last column containing the number following the old last number. The idea is that we want to compare these two numbers to check whether they form an interval (the transpose is due to how Matlab handels step 2):
a = [ a(:, 1)-1, a, a(:,2) + 1]';
We now have \[ a = \begin{bmatrix} 0 & 2 & 8\\ 1 & 3 & 9 \\ 2 & 6 & 13\\ 3 & 7 & 14 \end{bmatrix} \] Now we transform $a$ to a row vector throwing away the first and last element (we don't need them). This is why we had to transpose $a$ in the first step (as Matlab does this columnwise)
a = a(2:length(a(:))-1);
Now \[ a = \begin{bmatrix} 1 & 2 & 3 & 2 & 3 & 6 & 7 & 8 & 9 & 13 \end{bmatrix} \] if we split this into intervals there are legal ones like $[1,2]$ and illegal ones like $[3,2]$. The next two steps will find the first indices of the illegal ones and remove them afterwards.
I = find(a(1:length(a)-1) > a(2:length(a))); a([I, I + 1]) = [];
We got \[ a = \begin{bmatrix} 1 & 2 & 3 & 6 & 7 & 8 & 9 & 13 \end{bmatrix} \] Now we just have to reshape $a$ as a matrix again, undoing the transpose we did in the beginning
a = reshape(a, 2, length(a)/2)';
giving \[ a = \begin{bmatrix} 1 & 2 \\ 3 & 6 \\ 7 & 8 \\ 9 & 13 \end{bmatrix} \] I hope I understand correctly what you wanted to do, the program itself may not be the nicest solution, but it uses no for-loops directly :)