WLOG, let $ a<b<c<d $. The smallest sum is $ a+b=16 $. The second-smallest sum is $ a+c=19 $. The second-largest sum is $ b+d=22 $. The largest sum is $ c+d=25 $. In summary, \begin{align*}\tag{1} a+b&=16\\ \tag{2} a+c&=19\\ \tag{3} b+d&=22\\ \tag{4} c+d&=25 \end{align*} There are two sums left, $ a+d $ and $ b+c $. We will break this problem into two cases, the first case in which the first of the two sums is smaller than the second, and the second case in which the first of the two sums is larger than the second. In the first case \begin{align*} \tag{5} a+d&=20\\ \tag{6} b+c&=21 \end{align*} Adding Equations (1) and (6) and subtracting (2), we have $ (a+b)+(b+c)-(a+c)=16+21-19\Rightarrow b = 9 $. Plugging this value into Equation (1), we find that $ a+9=16 \Rightarrow a=7 $. Plugging the value of $ a $ into Equation (2), we find that $ 7+c=19 \Rightarrow c=12 $. Plugging the value of $ c $ into Equation (4), we find that $ 12+d=25 \Rightarrow d=13 $. Thus, the four integers are $ 7,9,12,13 $. In the second case, \begin{align*} \tag{7} b+c&=20\\ \tag{8} a+d&=21 \end{align*} Adding Equations (1) and (7) and subtracting Equation (2), we have $ (a+b)+(b+c)-(a+c)=16+20-19 \Rightarrow b=8.5 $. This case is impossible because $ b $ is defined to be an integer. Thus, the only solution is $ \boxed{7,9,12,13} $. Alternate solution: Again WLOG, assume $ a<b<c<d $. Then $ a+b=16 $ implies $ b \geq 9 $, and $ c+d=25 $ implies $ c \leq 12 $. Since $ a+b=16 $ and $ a+c=19 $, $ c-b=3 $, and so we must have $ b=9, c=12 $. Plugging back in, we get $ a=7, d=13 $, or $ a,b,c,d = \boxed{7,9,12,13} $