For knapsack, you can do that easily in polynomial time. Well, given a few minimal assumptions, like P!=NP; because otherwise there are no hard instances.
First, you start with an NP problem where virtually all instances are expected to be hard, like finding the pre-image to a given sha256 hash digest. Second, you sample a random instance in O(n). Third and last, you reduce this problem from the original sha256 inversion to knapsack. You can do this in polynomial time, because knapsack is NP complete.
Note for the pedantic: inverting sha256 is certainly in NP, but it's not expected to be NP complete.
Second note for the pedantic: because sha256's digest has a specific fixed size, you can technically solve any problems around it in constant time with a big lookup table. So you should replace sha256 in my example with any other problem that's expected to be hard on average.
