Summation of large inputs
Your approach takes \$O(n^2)\$ time. Since n can be as large as 100,000, you'll likely need something better. Some benchmarks (inputs are the numbers 1 to n shuffled, and the table shows times in seconds):
     n | original  AJNeufeld  AJ_no_islice  Kelly_Bundy 
-------+------------------------------------------------
  1000 |   0.12       0.07          0.06         0.07   
 10000 |  11.92       7.98          7.23         0.09   
 20000 |  50.82      32.83         29.79         0.11   
100000 |      -          -             -         0.18   

I didn't run the first three solutions on n=100000, they're too slow. AJ_no_islice is the same as AJNeufeld, except with a regular slice instead of islice to demonstrate that islice is not faster but slower.
My solution in a function that takes a list of numbers and returns the result (much better for testing, and for clarity):
def solve(numbers):
    incs = [0] * 100001
    for numerator, freq in Counter(numbers).items():
        for multiple in range(numerator, 100001, numerator):
            incs[multiple] += freq
    incs = list(accumulate(incs))
    return sum(incs[denominator] for denominator in numbers)

It's similar to the sieve of Eratosthenes, though I mark multiples of all numbers, not just of primes. Runtime is O(n log n).
Consider some numerator, let's say 3. Then for every denominator 3 or higher, it contributes 1 to the total. For every denominator 6 or higher, it contributes 2 to the total. And so on. As a table:
incs = [0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, ...]
So if you then consider some denominator, let's say 7, you simply look up incs[7] to see that it contributes 2 to the total (with the numerator 3).
Now instead consider some other numerator, let's say 4. Its table is:
incs = [0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, ...]
Together for numerators 3 and 4:
incs = [0, 0, 0, 1, 2, 2, 3, 3, 4, 5, 5, 5, 7, 7, 7, 8, 9, 9, ...]
How to efficiently compute that table? First only mark the increment spots, e.g., for denominator 3:
incs = [0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, ...]
And for denominators 3 and 4 together:
incs = [0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 2, 0, 0, 1, 1, 0, 1, 0]
And then just accumulate the prefix sums.
Full benchmark code (includes correctness checking):
from timeit import repeat
from random import choices, shuffle
from collections import Counter
from itertools import islice
from itertools import accumulate

def original(numbers):
    numbers = sorted(Counter(numbers).items())
    sigma = 0
    for i in range(len(numbers)):
        for j in range(i, len(numbers)):
            sigma += (numbers[j][0] // numbers[i][0]) * numbers[j][1] * numbers[i][1]
    return sigma

def AJNeufeld(numbers):
    numbers = sorted(Counter(numbers).items())
    sigma = sum((a_j // a_i) * a_j_count * a_i_count
                for i, (a_i, a_i_count) in enumerate(numbers)
                for a_j, a_j_count in islice(numbers, i, None))
    return sigma

def AJ_no_islice(numbers):
    numbers = sorted(Counter(numbers).items())
    sigma = sum((a_j // a_i) * a_j_count * a_i_count
                for i, (a_i, a_i_count) in enumerate(numbers)
                for a_j, a_j_count in numbers[i:])
    return sigma

def Kelly_Bundy(numbers):
    incs = [0] * 100001
    for numerator, freq in Counter(numbers).items():
        for multiple in range(numerator, 100001, numerator):
            incs[multiple] += freq
    incs = list(accumulate(incs))
    return sum(incs[denominator] for denominator in numbers)

funcs = original, AJNeufeld, AJ_no_islice, Kelly_Bundy

# Correctness
numbers = choices(range(1, 100001), k=1000)
expect = funcs[0](numbers)
for func in funcs:
    result = func(numbers)
    print(result == expect, func.__name__,)
print()

# Speed
names = [f' {func.__name__} ' for func in funcs]
print('     n |' + ''.join(names))
print('-------+' + '-' * sum(map(len, names)))
for n in 1000, 10000, 20000, 100000:
    print(f'{n:6} |', end='')
    numbers = list(range(1, n+1))
    shuffle(numbers)
    for func in funcs:
        if n &lt;= 20000 or func is Kelly_Bundy:
            t = min(repeat(lambda: func(numbers), number=1, repeat=1))
            t = '%.2f' % t
        else:
            t = '-'
        print(f'{t:&gt;{len(func.__name__)-1}}   ', end='')
    print()