Are the number of equations sufficient?

Least value = $a+b=1$

2nd least value= $a+c=11$

$3rd$ and $4th$ values are $21$ and $31$ of $b+c$ and $a+d$ , but in any order

So we have $2$ pairs of $4$ variable $4$ equation, i.e.

$a+b=1,a+c=11,b+c=21,a+d=31$

And

$a+b=1,a+c=11,b+c=31,a+d=21$

Solving we get $2$ possible values of d as $30.5$ and $35.5$ , which sum up to $\boxed {66}$ .

Note:

Here I have not considered $b+d$ and $c+d$ as they are the largest terms and sum of $4$ smallest terms is given.

And $a <b$ and $d>c$ implies that no comparison can be made between $a+d$ and $b+c$ . Either can be greater as adding the two mentioned inequalities can result in either $>,=,<$ sign