At each stage of the process, two numbers are chosen from the set. Call them a and b. They are replaced by the single number (a+b)/4. If we look at the number a and trace back where it came from, we see that it is either one of the original 1s, or it the result of combining two other numbers. If we continue tracing these numbers back, we see that a is in fact the outcome of an amalgamation process starting with a subset of the original n 1s. If this subset contains p 1s then a is the outcome of an amalgamation process starting with p 1s. Similarly b is the outcome of an amalgamation process starting with q 1s.
The idea is to prove by induction that the outcome of an amalgamation process starting with n 1s is at least 1/n. This is trivially true for n=1. Suppose as a (strong) inductive hypothesis that it is true for all numbers less than n.
Take an amalgamation process starting with n 1s, and consider the final stage of this process, in which the two remaining elements a and b are combined to form the outcome c=(a+b)/4. Since a is the outcome of an amalgamation process starting with p 1s, it follows from the inductive hypothesis that a≥1/p. Similarly, b≥1/q. Also, since this is the final operation in the amalgamation process, p+q=n.
Therefore c ≥ ((1/p) + (1/q))/4 = (p+q)/4pq. We want to show that this is ≥1/(p+q). Equivalently, we must show that (p+q)^2≥4pq. But this follows from the fact that (p+q)^2–(p-q)^2 = 4pq. That completes the inductive proof. All that remains is to put n=1994.