smc

Assume that $y-z=a,x-y=b,w-x=c.$ $w-z$ results in the greatest pairwise difference, and thus it is $9$. This means $a+b+c=9$. $a,b,c$ must be in the set $1,3,4,5,6$. The only way for 3 numbers in the set to add up to 9 is if they are $1,3,5$. $a+b$, and $b+c$ then must be the remaining two numbers which are $4$ and $6$. The ordering of $(a,b,c)$ must be either $(3,1,5)$ or $(5,1,3)$. $$\begin{array}{rl}z+(z+a)+(z+(a+b))+(z+(a+b+c))& =4z+a+(a+b)+9\\ 4z+a+(a+b)+9& =44\\ \text{if}a& =3\\ a+b& =4\\ 4z& =44-9-3-4\\ z& =7\\ w& =16\end{array}$$ $$\begin{array}{rl}\text{if}a& =5\\ a+b& =6\\ 4z& =44-9-5-6\\ z& =6\\ w& =15\end{array}$$ The sum of the two w's is $15+16=31$ $\boxed{31}$