Question 1041570
{{{(A+B+C+D)/4=12}}}
{{{A+B+C+D=48}}}
Since they are unique, positive, and even, the smallest three integers you could have would be 2,4,6. 
Using those you can find what the maximum remaining positive even integer could be,
{{{2+4+6+D=48}}}
{{{D=36}}}