For such problems (esp when you notice that there is a general pattern), some ideas are to find a recurrence relation, create something telescoping (or treat it as a generating function).
We'd use these ideas here.
Notice that $ \left(\frac{1}{n-m} - \frac{1}{n}\right) { n - m \choose m } = \frac{m}{ n (n-m) } { n - m \choose m } = \frac{1}{n} {n-m-1 \choose m-1}$, or that
$$ \frac{ 1 } { n-m } { n-m \choose m } = \frac{1}{n} \left[ { n - m \choose m } + { n - m - 1 \choose m- 1 } \right]. $$
This is a good substitution, as it gets rid of the pesky $ \frac{1}{n-k}$ which makes recurrence hard, and also gives us a $\frac{1}{1991}$ on the RHS.
Thus, the goal is to determine $ \sum_{k=0}^{995 } (-1)^k \left[ {1991-k\choose k} + { 1991 - k - 1 \choose k - 1 } \right] $. (We will show that it equals to 1, and thus the desired sum is $\frac{1}{1991}.$)
Let $ S_n = \sum_{k=0}^{ \lfloor \frac{n}{2} \rfloor} (-1)^k { n-k \choose k } $.
Notice that ${n-k \choose k } = { n-k - 1 \choose k } + { n-k - 1 \choose k - 1 } $, so
$ S_n = \sum_{k=0}^{\lfloor \frac{n+1}{2} \rfloor} (-1)^k { n - k + 1 \choose k } \\
= \sum_{k=0}^{\lfloor \frac{n+1}{2} \rfloor} (-1)^k \left[ {n-k \choose k } + {n-k \choose k - 1 } \right] \\
= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k {n-k \choose k } + \sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} (-1)^k { n-k \choose k } \\
= S_{n} - S_{n-1}. $
(Take care in checking the indices, and remember those ${n \choose m } = 0 $ when $m > n $.)
Using this recurrence relation, and calculating some initial values, we get $S_n = 1 , 0, -1, -1, 0, 1, 1, 0, -1, \ldots$, which has period 6.
We thus want to determine $S_{1991} - S_{1990} = 0 - (-1) = 1$.
Notes
I do wish there was a combinatorial argument here. For example, $S_n$ has an immediate interpretation as the difference between the even and odd permutations $p$ such that $|p(i) - i | \leq 1$. (IE Out of the first $n$ integers, there are ${n-k \choose k }$ ways to pick k pairs of consecutive integers (for a total of 2k). The perumatation which switches these pairs and keep the rest fixed has parity $k$.) However, I don't see an obvious way to show that this difference is $1, 0, -1, -1, 0, 1, \ldots $.
WhatsUp's conclusion that about the value of $s_n$ also follows from the above.