0-1 knapSnack problem with negetive numbers

Given an array of positive and negative numbers, find if we can choose some of them and sum them in a way that the final sum is zero. (any non zero number will do).

We have weights of them and values.

I believe this is a version of the 0-1 knapsack problem.
the sample input and output:

4 - weight - percent
+ 50 30
+ 80 1
- 20 30
- 30 30

yes

This is the code I wrote for it, but somehow I can't figure out why this won't work:
I'm getting the output 0 for it (and I should be getting 1000, choosing the first, third and fourth one).
Is there any way to solve the knapsack problem with negative values?

def knapSack(W , wt , val , n): 
    if n == 0 : 
        return 0
    if (wt[n-1] > W): 
        return knapSack(W , wt , val , n-1) 
    else: 
        return max(val[n-1] + knapSack(W-wt[n-1] , wt , val , n-1), knapSack(W , wt , val , n-1)) 
  
val = [80, 50, -20, -30] 
wt = [ 1,30, 30, 30] 
W = 0
n = len(val) 
print(knapSack(W, wt, val, n)) 

Anyone has any idea how should I change this in order to be working??

This is by no ways efficient but it seemed to be the simplest method of seeing if any combination of acids and bases would cancel out. (I did not use the knapsack approach)

For each (the acids and the bases) create an array that contains their amount. In the case of your example these arrays would be - acid = [1500, 80] and base = [600, 900]. Getting these arrays depends on your input format.

once you have these arrays you can do the following -

def SumOfSubsets(arr): 
	Allsums = []
	for number in arr:
		Placeholder = Allsums[:]
		Allsums.append(number)
		for sum in Placeholder:
			Allsums.append(sum + number)
  	
	return(Allsums)

AcidList = SumOfSubsets(acid)
BaseList = SumOfSubsets(base)

for acids in AcidList:
    if acids in BaseList:
        print("can be neutralized")

*Note that by using this approach you do not know what combinations of acids and bases are needed to achieve the neutralization. You can get this combination by using another function if needed.