(Looks like @apm got the same answer while I was writing, he was the first one to publish this particular answer.)
The first number can be as small as
Because of the restrictions, the 13th number cannot be more than $13 -\epsilon$.
Because of the "ordered numbers" restriction, all the earlier numbers must be smaller than that, for example $13-2\epsilon$, $13-3\epsilon$ and so on. (The exact choice of the numbers isn't important, as long as they are "very close" to 13)
The second smallest number must make for an integer sum with the larger numbers, so its non-integer part must sum up to an integer with the epsilons (all $-66$ of them) from earlier. Therefore the second number cannot be larger than $12+66\epsilon$, so it can contribute only $12$ to the total, while the numbers 3 to 13 will be able to contribute a whole $13$. Therefore the maximum sum from the the last 12 numbers is
$11 \times 13 + 12 = 155$
and therefore, the sum can add up to 13 when
the first number is $13 - 155 = -142$.
If we choose $\epsilon = 0.01$, the numbers become