Introduce every group, then start catching.
Learned myself a new mathematical concept.
This question is a variation on the Coupon Collector's problem: "Given n coupons, how many coupons do you expect you need to draw with replacement before having drawn each coupon at least once?" the simplified solution being:
E(T) = n * Hn - k
Where E(T) is the estimated number of draws required, n is the number of unique coupons there are, k is the number of unique coupons you've already drawn before starting the trial, and Hn - k being the (n - k)th harmonic number.
So, if we introduce the smallest group of Unown first, catch them all, then introduce the next smallest and so forth, the expected number of encounters looks like this:
E(T) = 3 * H3 + 8 * H5 + 15 * H7 + 26 * H11 = 141.176
Contrasting introducing the largest group first and going smaller and smaller, we get:
E(T) = 11 * H11 + 18 * H7 + 23 * H5 + 26 * H3 = 180.073
Finally, let's look at releasing all the Unown at once and then starting the hunt:
E(T) = 26 * H26 = 100.215
So as it would turn out, releasing all the Unown at once is far and away the best solution.
I wasn't factoring in Unown Z's reduced spawn rate in the calculation, so it may be worthwhile (for peace of mind if nothing else) to get X, Y and Z first anyway.