(Analysis by Nick Wu)

If we sort the diamonds by size and pick the sizes of the two smallest diamonds that will appear in each case, we can use binary search to figure out, in $O(\log N)$ time, the size of each case. However, there are too many diamonds for us to brute force try all possible pairs.

There exists an optimal arrangement of diamonds such that all the diamonds in one case are strictly smaller than all the diamonds in the other case. This can be seen by taking an optimal arrangement which may not satisfy this constraint, taking the smallest diamond and putting all diamonds that can be presented with it on one case, and putting all the other diamonds in the other case.

Imagine that we know beforehand what the maximum number of diamonds we can stick in a case given that all the diamonds have size at least K, as well as the maximum number of diamonds we can stick in a case given that all the diamonds have size at most K. We can then loop over all possible values of K.

To actually compute the desired values, we can precompute for each diamond size how many diamonds we can have in a case with a given minimally/maximally-sized diamond. We can then loop over all of these precomputed values and save the maximum number of diamonds given that all diamonds are at least or at most a given size.

Here is my Java solution.

import java.io.*;
import java.util.*;
public class diamondS {
public static void main(String[] args) throws IOException {
PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("diamond.out")));
int n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());
int[] list = new int[n];
for(int i = 0; i < n; i++) {
}
Arrays.sort(list);
// leftmostIndex[i] stores the index of the smallest diamond that can be included given that
// the largest diamond in the case has size list[i].
int[] leftmostIndex = getLeftmost(list, k);
// leftSize[i] stores the maximum number of diamonds given that all diamonds have size at most list[i].
int[] leftSize = new int[n];
for(int i = 0; i < n; i++) {
leftSize[i] = i - leftmostIndex[i] + 1;
if(i > 0) {
leftSize[i] = Math.max(leftSize[i], leftSize[i-1]);
}
}
// rightmostIndex[i] stores the index of the smallest diamond that can be included given that
// the smallest diamond in the case has size list[i].
int[] rightmostIndex = getRightmost(list, k);
// leftSize[i] stores the maximum number of diamonds given that all diamonds have size at least list[i].
int[] rightSize = new int[n];
for(int i = n-1; i >= 0; i--) {
rightSize[i] = rightmostIndex[i] - i + 1;
if(i < n-1) {
rightSize[i] = Math.max(rightSize[i], rightSize[i+1]);
}
}
int ret = 0;
for(int i = 0; i < n-1; i++) {
ret = Math.max(ret, leftSize[i] + rightSize[i+1]);
}
pw.println(ret);
pw.close();
}

public static int[] getRightmost(int[] list, int k) {
int[] ret = new int[list.length];
int j = list.length-1;
for(int i = list.length-1; i >= 0; i--) {
while(j >= 0 && list[j] - list[i] > k) {
j--;
}
ret[i] = j;
}
return ret;
}

public static int[] getLeftmost(int[] list, int k) {
int[] ret = new int[list.length];
int j = 0;
for(int i = 0; i < list.length; i++) {
while(j < list.length && list[i] - list[j] > k) {
j++;
}
ret[i] = j;
}
return ret;
}

}