sequoia-

Although gsaha may be right for 1.91+, that approach is not needed for 1.92.

In fact, you almost had it. Instead of:

{sum(q1.NAOK,q2.NAOK,q3.NAOK,q4.NAOK)}

use

{sum(q1.value,q2.value,q3.value,q4.value)}

You could also use

{sum(q1.valueNAOK,q2.valueNAOK,q3.valueNAOK,q4.valueNAOK)}

but, since q1-q4 are always relevant, there is no need for the NAOK suffix

The .value suffix retrieves the assessment value for the answer. When you were using {sum(q1,q2,q3,q4)}, it was retrieving the answer id, so say you answered A1,B2,C3,D1, {sum{q1,q2,q3,q4)} would be 0, since the integer value of "A1" is 0, whereas {sum(q1.value,q2.value,q3.value,q4.value)} would be the equivalent of {sum(5,10,9,1)}, and returns the assessment value of 35.