It's NP complete, but you may have a chance in many practical cases, especially if you need to produce a lot of numbers in a not very large range. I'll call the numbers you want to produce "desired numbers", and assume that 1 is not one of the "desired numbers".
An important observation is that in any chain, performing an addition that produces one of the desired numbers is optimal. Therefore, we need only count the additions that don't produce one of the desired numbers (and if you want to produce 100 numbers from 1 to 200 those additions might be very few).
So instead of the initial set { 1 }, we add { 2 } if it is one of the desired numbers, then add 3 or 4 if they are among the desired numbers, and so on. If you wanted { 2, 3, 5, 6, 10, 19, 20, 21, 50 } then you start with the initial set { 1, 2 = 1+1, 3 = 2+1, 5 = 3+2, 6 = 5+1, 10 = 5+5, 20 = 10+10, 21 = 20+1 } and you only need to find 19 and 50.
Then each move consists of adding the sum of two numbers, and then adding all of the desired numbers that can now be produced. In the example, one move would be adding 16 and 19, another move adding 40 and 50. All this really cuts down on the number of possibilities you need to try out.
Some more ways to cut down on the possibilities: We can require that the non-desired numbers are produced in increasing order. So if you produced 2 and 4, you cannot produce the number 3 anymore. Or if you produced 2, 3, 5, you cannot produce the number 4 anymore - if you wanted it, you should have produced it earlier.
Obviously you don't ever produce a number larger than the highest desired number. Or higher than the highest desired number that wasn't produced yet. And all numbers greater than the highest desired number that wasn't produced yet must have been used producing one of the higher desired numbers.
And you cannot ever produce a number larger than the smallest desired number not found yet.
Apart from that, I cannot think of any "elegant" algorithm. I'd try all the possibilities, producing large numbers first, and of course stopping when you can prove you are not going to find a solution than the best so far.
Your example: [16, 30, 36, 40].
First attempt, always adding the largest number possible according to the rules: 2, 4, 8+16, 24, 28+30, 34+36+40. That's six moves, so you can try to find a better solution using five or fewer moves using backtracking. (Note that in move 4 and 5 we couldn't produce a number > 30 according to the rules).
Producing 26 or 25 in step 5 doesn't help. In step 4, we could produce 20+36+40, 18+36+40, 17, 12, 10, 9. 20+36+40 can be followed by 28+30, so we found a better solution using five moves only (2, 4, 8+16, 20+36+40, 28+30). Now the goal is to find a chain with four or fewer extra additions.
2, 4, 8+16 doesn't give success with four extra additions. We can try 2, 4, 6 or 2, 4, 5 or 2, 3, 6 or 2, 3, 5 or 2, 3, 4 which all don't solve the problem in four moves. So (2, 4, 8+16, 20+36+40, 28+30) with five moves plus four moves adding the four desired numbers is an optimal solution.