I've calcuated U(4,13) = 11/6 (using Veit's notation) using CPLEX (industrial strength MIP) with the MIP that Veit suggested.
i can now determine T(4, n) in general. recall that my normalization is that T(m, n) equals the maximum c such that there is a partition of 1 with all parts >= c , which is simultaneously a refinement of 1/m + 1/m + ... + 1/m = 1 and also of 1/n + 1/n + ... + 1/n = 1 . T(4, 4n) = 1/4n , with unique partition 4n * 1/(4n) . this is obvious. T(4, 4n+2) = 1/(8n+4) . 4 * [(2n+1) * 1/(8n+4)] <---> (4n+2) * [1/(8n+4) + 1/(8n+4)] achieves this bound. moreover, at least one 1/(4n+2) must be split, so it has a part at most 1/(8n+4) . the partition is not unique; at least two 1/(4n+2) 's must split into 1/(8n+4) + 1/(8n+4) . the others either split in the same way, or do not split. T(4, 2n+1) = (2n-1)/(4n(2n+1)) if n >= 1 . the case n = 1 is T(4, 3) = T(3, 4) = 1/12 , which is easy to handle separately. if n > 1 , there is a unique optimal partition, which is 2 * [n * (2n-1)/(4n(2n+1)) + 1/(4n+2)] + 2 * [n * 1/(4n)] <---> 2n * [(2n-1)/(4n(2n+1)) + 1/(4n)] + [1/(4n+2) + 1/(4n+2)] suppose that 1/4 splits as 1/(2n+1) + 1/(2n+1) + ... + 1/(2n+1) + ( p smaller parts ) , where there are k > 0 parts of size 1/(2n+1) . if p >= n - 2k + 1 , then the smallest of the parts is at most (1/4 - k/(2n+1))/p <= ((2n-4k+1)/(4(2n+1)))/(n - 2k + 1) = (2 - 1/(n - 2k + 1))/(4(2n+1)) < (2 - 1/n)/(4(2n+1)) = (2n-1)/(4n(2n+1)) , so this part is too small. on the other hand, if p <= n - 2k , then the largest of the parts not equal to 1/(2n+1) is at least (1/4 - k/(2n+1))/p . since this occurs as a part of 1/(2n+1) , it has another part at most 1/(2n+1) - (1/4 - k/(2n+1))/p <= 1/(2n+1) - (1/4 - k/(2n+1))/(n - 2k) = (2 - 1/(n - 2k))/(4n(2n+1)) < (2 - 1/n)/(4(2n+1)) = (2n-1)/(4n(2n+1)) , so again there is a part which is too small. this shows that no 1/4 can contain any parts of size 1/(2n+1) , and therefore every 1/(2n+1) must split. now suppose that 1/4 splits into p parts, which we know must all be less than 1/(2n+1) . if p >= n+2 , then the smallest of these parts is <= 1/(4(n+2)) < (2n-1)/(4n(2n+1)) , so too small. (here we need n > 1 .) if p <= n-1 , then the largest of these parts is at least 1/(4(n-1)) , and since it occurs as part of 1/(2n+1) , there must be another part of size <= 1/(2n+1) - 1/(4(n-1)) < (2n-1)/(4n(2n+1)) , which is again too small. therefore, each 1/4 splits into either n or n+1 parts. if 1/(2n+1) splits into 3 or more parts, then some part has size <= 1/(3(2n+1)) , which is less than (2n-1)/(4n(2n+1)) if n > 1 . thus each 1/(2n+1) splits into exactly 2 parts, so there are 4n+2 parts in total. now two of the 1/4 's must split into n parts, and the other two must split into n+1 parts. moreover, the largest part in either of the 1/4 's that split into n parts has size >= 1/(4n) . since it occurs as part of 1/(2n+1) , the other part has size <= 1/(2n+1) - 1/(4n) = (2n-1)/(4n(2n+1)) . thus, to have optimality, both of the 1/4 's that split into 1/(4n) + 1/(4n) + ... + 1/(4n) , and now n of the 1/(2n+1) 's must split as (2n-1)/(4n(2n+1)) + 1/(4n) . finally, there are 2n parts of the minimal size (2n-1)/(4n(2n+1)) , and neither of the remaining 1/4 's can contain n+1 of them , so both contain n of them, and thus split as n * (2n-1)/(4n(2n+1)) + 1/(4n+2) , which determines the partition. T(1, n) and T(2, n) are very easy to determine in general. it looks like there is probably also pattern in T(3, n) . T(6, n) might also be possible to determine in general. mike