(11) The correct answer is $(D)$.

Such an isosceles triangle indeed has a base equal to 2018 and two equal sides, denoted as $n$, which are shorter. It must then be that $0<n<2018$ and, by the triangle inequality, $2018<2n$. It follows that the possible measures of the two equal sides vary among the integers strictly between 0 and 1009: the possible triangles are thus 1008.

Question proposed by Carmelo Di Stefano.
11. How many different isosceles triangles, with all sides of integer lengths, have one side of length 2018 that is greater than the other two?
(A) 1007
(B) 1006
(C) 1010
(D) 1008
(E) 1011