Java - How to Solve this 2D Array Hour Glass?

Remove the if (arr[arr_i][arr_j] > 0) statement. It prevents finding the answer at row 1, column 0, because that cell is 0.

Comments for other improvements to your code:

• What if the best hourglass sum is -4? You should initialize tmp_sum to Integer.MIN_VALUE. And name it maxSum, to better describe it's purpose.

• You shouldn't define sum outside the loop. Declare it when it is first assigned, then you don't have to reset it to 0 afterwards.

• Your iterators should be just i and j. Those are standard names for integer iterators, and keeps code ... cleaner.
  If you prefer longer names, use row and col, since that is what they represent.

• You don't need parenthesis around the array lookups.

• For clarity, I formatted the code below to show the hourglass shape in the array lookups.

Scanner in = new Scanner(System.in);
int arr[][] = new int[6][6];
for (int i = 0; i < 6; i++){
    for (int j = 0; j < 6; j++){
        arr[i][j] = in.nextInt();
    }
}

int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
        int sum = arr[i    ][j] + arr[i    ][j + 1] + arr[i    ][j + 2]
                                + arr[i + 1][j + 1]
                + arr[i + 2][j] + arr[i + 2][j + 1] + arr[i + 2][j + 2];
        if (maxSum < sum) {
            maxSum = sum;
        }
    }
}
System.out.println(maxSum);