Other than n = 1, it is easy to see that too small n's do not have such sets:
n = 2: sum is 6, which does not have two-squares decomposition.
n = 3: sum is 21, whose only three-squares decomposition is 16 + 4 + 1, but it is impossible because 6 + 5 + 4 < 16, or alternatively, 1 + 2 + 3 > 1 + 4.
n = 4: sum is 55, which has only three distinct four-squares decomposition (25 + 25 + 4 + 1), (36 + 9 + 9 + 1), (49 + 4 + 1 + 1), none of which can solve the problem due to a similar argument.
The first n that has a solution in both parts is
n = 5
which can be solved as follows:
1 = 1
9 = 3 + 6
25 = 2 + 10 + 13
36 = 4 + 5 + 12 + 15
49 = 7 + 8 + 9 + 11 + 14
Given this, I conjecture that
all numbers n >= 5 permit such set partitions (even with the restriction of distinct squares), with at most a small finite number of exceptions
because
1) It is known that every positive integer has one or more four-square decompositions, and n-square decompositions (with n >= 5) will give more and more freedom in the choice of square numbers in the decomposition.
2) Sum of first T(n) numbers is quartic in n, which grows asymptotically faster than the sum of first n squares (which is cubic in n), supporting 1) further.
3) For any given n-square decomposition, having more numbers to choose from for each set means it is more likely to find a set partition fitting the given sums.