Existence 1

Assume that magic squares of order 2 exist. Let the grid below represent the magic square of order 2: $\text{A B}$ $\text{C D}$ The magic sum for this square is $M_2=\frac{2}{2}(2^2+1)=5$ . Thus: $A+B=5$ $C+D=5$ $A+C=5$ $B+D=5$ $A+D=5$ $B+C=5$ Subtracting the third from the first equation yields $B-C=0$ , or $B=C$ . Subtracting the sixth from the fourth equation yields $D-C=0$ , or $D=C$ . Finally, subtracting the fifth from the fourth equation yields $B-A=0$ , or $B=A$ . Combining all the results above, $A=B=C=D$ a contradiction to the fact that the numbers to be put into the magic square must be the distinct integers 1, 2, 3, 4. Thus our assumption is wrong. Therefore, there doesn't exist a magic square of order 2.

The answer is $\boxed{0}$ .