Let the integers in increasing order be:


\(a_1<a_2<a_3<a_4<a_5<50\)


Their sum is \(6 \times 20=120\) , so


\(a_1+a_2+a_3+a_4+a_5=70\)


We want to maximize \(a_3+a_4\) .
To make \(a_3+a_4\) as large as possible, make \(a_1, a_2, a_5\) as small as possible:

Smallest possible: \(a_1=1, a_2=2\)
Since \(a_5>a_4\) , the smallest possible \(a_5 \) is \(a_4+1\)


Then:

\( 1+2+a_3+a_4+\left(a_4+1\right) \leq 70 \\
a_3+2 a_4 \leq 66\)


Let \(S=a_3+a_4\) . Since \(a_3<a_4 \) we have \(S<2 a_4\) , so \(a_4>\frac{S}{2}\) .
Also from \(a_3+2 a_4 \leq 66\) :


\(S+a_4 \leq 66\)


\(a_4 \leq 66-S\)


So:


\(\frac{S}{2}<66-S \\
\frac{3 S}{2}<66 \\
S<44\)


Thus the greatest possible integer S is 43 .

Check if S=43 works:
Take \(a_3=20, a_4=23\) , then:


\(a_5=70-1-2-20-23=24\)


So the list is:


1,2,20,23,24,50


Sum =120 , average =20 , and \(a_3+a_4=43\) .

Therefore, the greatest possible sum is:



Answer: (E) 43