Suppose that Dewa's four numbers are $w, x, y, z$. The averages of the four possible groups of three of these are $\frac{w+x+y}{3}, \frac{w+x+z}{3}, \frac{w+y+z}{3}, \frac{x+y+z}{3}$. These averages are equal to $32,39,40,44$, in some order. The sums of the groups of three are equal to 3 times the averages, so are $96,117,120,132$, in some order. In other words, $w+x+y, w+x+z, w+y+z, x+y+z$ are equal to $96,117,120,132$ in some order. Therefore, $(w+x+y)+(w+x+z)+(w+y+z)+(x+y+z)=96+117+120+132$ and so $3 w+3 x+3 y+3 z=465$ which gives $w+x+y+z=155$. Since the sum of the four numbers is 155 and the sums of groups of 3 are $96,117,120,132$, then the four numbers are $155-96=59 \quad 155-117=38 \quad 155-120=35 \quad 155-132=23$ and so the largest number is 59.