Find all the 4 digit vampire numbers

Or you can use a property of vampire numbers described on this page (linked from Wikipedia) :

An important theoretical result found by Pete Hartley:

      If x·y is a vampire number then x·y == x+y (mod 9) 

Proof: Let mod be the binary modulo operator and d(x) the sum of the decimal
digits of x. It is well-known that d(x) mod 9 = x mod 9, for all x.
Assume x·y is a vampire. Then it contains the same digits as x and y,
and in particular d(x·y) = d(x)+d(y). This leads to:
(x·y) mod 9 = d(x·y) mod 9 = (d(x)+d(y)) mod 9 = (d(x) mod 9 + d(y) mod 9) mod 9
= (x mod 9 + y mod 9) mod 9 = (x+y) mod 9

The solutions to the congruence are (x mod 9, y mod 9) in {(0,0),
(2,2), (3,6), (5,8), (6,3), (8,5)}

So your code could look like this :

for(int i=18; i<100; i=i+9){          // 18 is the first multiple of 9 greater than 10
   for(int j=i; j<100; j=j+9){        // Start at i because as @sh1 said it's useless to check both x*y and y*x
       checkVampire(i,j);
   }
}

for(int i=11; i<100; i=i+9){          // 11 is the first number greater than 10 which is = 2 mod 9
   for(int j=i; j<100; j=j+9){
       checkVampire(i,j);
   }
}

for(int i=12; i<100; i=i+9){
   for(int j=i+3; j<100; j=j+9){
       checkVampire(i,j);
   }
}

for(int i=14; i<100; i=i+9){
   for(int j=i+3; j<100; j=j+9){
       checkVampire(i,j);
   }
}

// We don't do the last 2 loops, again for symmetry reasons

Since they are 40 elements in each of the sets like {(x mod 9, y mod 9) = (0,0); 10 <= x <= y <= 100}, you only do 4*40 = 160 iterations, when a brute-force gives you 10ˆ4 iterations. You can do even less operations if you take into account the >= 1000 constraint, for instance you can avoid checking if j < 1000/i.

Now you can easily scale up to find vampires with more than 4 digits =)