Thijs Laarhoven has already pointed out the sign errors in your calculations. As for the original question ("How do I know what sign it is?"), you just have to decide on some sign convention: if the result, reduced modulo $10^n$, is less than some threshold $M$, then you should interpret it as positive, otherwise as negative.
One natural choice of threshold is $M = 10^n/2$, which is equivalent to the rule suggested by Peter Taylor: if the $n$-th digit is $5$ or greater, the number is negative. However, pretty much any other choice of $0 < M \le 10^n$ could be used as well. ($M=10^n$ corresponds to unsigned arithmetic modulo $10^n$.) In any case, you will always experience arithmetic overflow whenever the result becomes greater than or equal to $M$ or less than $M-10^n$.
This is exactly the same situation as in base 2; the convention that numbers with the highest bit set are considered negative (i.e. $M = 2^n/2 = 2^{n-1}$) is just one possibility out of many. One advantage of this particular convention, and of the similar convention for base 10 suggested above, is that they ensure that negation consistently flips the sign of almost all non-zero numbers; the sole exception in both cases is $M \equiv -M$, which is congruent to its own negative.
This is an unavoidable consequence of $b$'s complement representation for even $b$: since there are an even number of possible numbers, and since one of them must represent zero, that leaves an odd number of numbers to be divided among the positives and negatives.