Stick with Numpy array and use its sum() method:

>>> arr = np.array([[1,2,3,5,4,3], 
          [5,7,2,4,6,7],
          [3,6,2,4,5,9]])
>>> arr.sum(axis=0)
array([ 9, 15,  7, 13, 15, 19])

Of course you can do it with Python lists as well but it is going to be slow:

>>> lst = [[1,2,3,5,4,3], 
          [5,7,2,4,6,7],
          [3,6,2,4,5,9]]
>>> map(sum, zip(*lst))
[9, 15, 7, 13, 15, 19]

The zip function is useful here, used with a list comprehension.

[x + y for x, y in zip(first, second)]

If you have a list of lists (instead of just two lists):

lists_of_lists = [[1, 2, 3], [4, 5, 6]]
[sum(x) for x in zip(*lists_of_lists)]
# -> [5, 7, 9]

You can use numpy.

import numpy as np

a = np.array(arrayOne)
b = np.array(arrayTwo)
max = max(a + b)
print(max)

print(df.loc[df['year'].isin((1851,1852))]["occur"].sum())

Or:

 print(df.loc[df.year.isin((1851,1852))].occur.sum())

For a range of dates creating a list of ranges seems more efficient than using &:

df.loc[df.year.isin(range(s1, s2+1))].occur.sum()

Well, all other answers are awesome for adding 2 numbers (list of digits).
But in case you want to create a program which can deal with any number of 'numbers',

Here's what you can do...

def addNums(lst1, lst2, *args):
    numsIters = [iter(num[::-1]) for num in [lst1, lst2] + list(args)]  # make the iterators for each list
    carry, final = 0, []                                                # Initially carry is 0, 'final' will store the result
    
    while True:
        nums = [next(num, None) for num in numsIters]                   # for every num in numIters, get the next element if exists, else None
        if all(nxt is None for nxt in nums): break                      # If all numIters returned None, it means all numbers have exhausted, hence break from the loop
        nums = [(0 if num is None else num) for num in nums]            # Convert all 'None' to '0'
        digit = sum(nums) + carry                                       # Sum up all digits and carry
        final.append(digit % 10)                                        # Insert the 'ones' digit of result into final list
        carry = digit // 10                                             # get the 'tens' digit and update it to carry

    if carry: final.append(carry)                                       # If carry is non-zero, insert it
    return final[::-1]                                                  # return the fully generated final list

print(addNums([6, 9, 8], [5, 9, 2]))                                    # [1, 2, 9, 0]
print(addNums([7, 6, 9, 8, 8], [5, 9, 2], [3, 5, 1, 7, 4]))             # [1, 1, 2, 7, 5, 4]

Hope that makes sense!