Just some observations.

• Any integer natural number may be decomposed into a product of powers of primes: $n == do { my $prod = 1; $prod *= $p[$_]**$k[$_] for 0..$#p; $prod } where @p are prime and @k are positive integers.
• All proper divisors of $n are products of the form do { my $prod = 1; $prod *= $p[$_]**int rand($k[$_]+1) for 0..$#p; $prod } , and all such products are divisors.
• There are therefore factorial(sum( @k)) proper divisors of $n. ^*
• If $n is larger than the sum of all its divisors, no subset can sum to it either.
• If $n is odd, an odd number of divisors will be in a sum which equals $n. If $n is even, an even number of the odd divisors will be in the sum.

You are right that this is a hard problem. The factorization of $n alone is an age-of-the-universe problem for large $n. Math::Pari is able to factor numbers of up to 60 digits fairly quickly. It is fast for less than 50 digits.

^* Oops, that is not quite right. If an element of @k is greater than 1, there is a binomial coefficient of ways to forms a power of that prime. All those are equivalent, yielding the same factor. That makes the factorial function I leaped at significantly larger than the actual number of distinct factors. The correct number of distinct divisors of $n is do { $prod = 1; $prod *= ($_ + 1) for @k; $prod } . Thanks to OverlordQ for pointing me to the right calculation.

After Compline,
Zaxo