Full solution (assuming cards aren't removed). My write-up assumes familiarity with the symmetric group. I also don't know a proof that this is the theoretically best-possible solution.
The participants agree on an explicit ordering on the envelopes and cards (equivalently, the envelopes and the participants) ahead of time. The strategy will be: participant number j waltzes straight to the envelope marked j and looks for her card. If she finds card k in envelope j, she opens envelope k next. She proceeds in this way until she finds her card (or loses the game for the team).
That's it. The reason the odds that this succeeds are approximately 30% is as follows: we have a permutation f of the set {1, 2, ..., 200} given by the rule "if envelope number j contains card number k, then f(j) = k." Notice that the participants succeed using the given strategy if and only if this permutation does not contain a cycle of length greater than 100, since the strategy just has the participant loop in order through the cycle containing her number. So we just need to count these permutations. If a permutation in S_{2k} has a cycle of length greater than k, then it only has one such cycle. So the number of permutations of length m, where m > k, is given by the following formula (explained below):
(2k choose m)(m!/m)(2k-m)!
(Explanation of formula: we choose m elements to be "in" the cycle. We can order them in m! different ways, but two orderings are the same cycle if they differ by the n trivial reorderings which "cycle" the elements (couldn't think of a better verb there, sorry), so we divide by n. The (2k - m) elements that are "out" of the cycle can be grouped in any permutation in S_{2k - m}.
This formula simplifies to the much nicer looking
(2k)!/m.
So the probability that a permutation contains an m-cycle (i.e. is bad for our participants) is just 1/m, since there are (2k)! total permutations.
Thus our participants have probability 1/(k+1) + 1/(k+2) + ... + 1/(2k) of failing. This is a right-hand Riemann sum approximating the integral from k to 2k of dx/x, so is close to log(2k) - log(k) = log(2), which is where the 30% comes from (there are other ways to connect this sum to log(2) as well).