Don,

There isn't a perfect solution, but there is a pretty good one. The arithmetic does not work out to have every pair of people together at least once: there are 66 possible pairs and the schedule only has room for 60 pairs to be together (3 per boat per day).

In the schedule described below, each person is in each boat at least once, and every pair of people are together exactly once apart from very few exceptions. The people are 1, 2, ..., 9, a, b, c. There are 9 pairs missing, and three who are together twice. The missing pairs are 4 with each of 1, 2, 3; 8 with each of 5 6 7; and 9 with each of a b c. The only pairs together twice are 4 8, 4 9, and 8 9.

The boats are ordered left to right.

Time 1: 1 5 9; 2 6 a; 3 7 b; 4 8 c
Time 2: 3 8 a; 4 7 9; 1 6 c; 2 5 b
Time 3: 2 7 c; 1 8 b; 5 4 a; 3 6 9
Time 4: 4 6 b; 3 5 c; 2 8 9; 1 7 a
Time 5: 1 2 3; 5 6 7; a b c; 4 8 9

Hope this works for you.

--Victoria