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DanielLC comments on Probability puzzle: Coins in envelopes - Less Wrong Discussion
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I notice solving problem 2 also solves problem 1.
The solution to two:
sum(f(k)) = sum of f(k) for each value of k from 0 to m inclusive
1/2 = sum(k^2/m)/sum(k)-1/m
= (m(m+1)(2m+1)/6)/(m^2(m+1)/2)-1/m
= (2m+1)/3m-1/m
= (2m+1)/3m-3/3m
= (2m-2)/3m
= 1/2
4m-4 = 3m
m = 4
Four envelopes.