Lemma Time!

$S=\sum_{i=1}^{19} \frac{(-1)^{r-1} \times r}{20 \choose r}$

Replacing $r$ by $n-r$ and adding we get

$S=20\sum_{i=1}^{9} \frac{(-1)^{r-1}}{20 \choose r}-\frac{10}{20 \choose 10}$

Now

$\frac{1}{n \choose r} = \frac{n+1}{n+2}(\frac{1}{{n+1} \choose {r+1}}+\frac{1}{{n+1} \choose r})$

Splitting each term we get

$S=20 \times \frac{21}{22} (\frac{1}{21 \choose 1}+\frac{1}{21 \choose 2}-\frac{1}{21 \choose 2}-\frac{1}{21 \choose 3}.........+\frac{1}{21 \choose 10})-\frac{10}{20 \choose 10}$

Consecutive terms will get cancelled

$S=20 \times \frac{21}{22}(\frac{1}{21 \choose 1}+\frac{1}{21 \choose 10})-20 \times \frac{21}{22} \times \frac{1}{21 \choose 10}$

Therefore

$S=\boxed{\frac{10}{11}}$

For any positive even integer $n$ ,

$\displaystyle\sum_{r=1}^{n-1}\frac{(-1)^{r-1}r}{{n \choose r}}=\sum_{r=1}^{\frac{n}{2}-1}\frac{(-1)^{r-1}n}{{n \choose r}}+\frac{(-1)^{\frac{n}{2}-1}\frac{n}{2}}{{n \choose \frac{n}{2}}}=-\frac{n}{2}\sum_{r=1}^{n-1}\frac{(-1)^{r}}{{n \choose r}}$

$=\displaystyle-\frac{n}{2}\sum_{r=0}^{n}\frac{(-1)^{r}}{{n \choose r}}+n=-\frac{n}{2}(1+(-1)^{n})\left(\frac{n+1}{n+2}\right)+n$

$=\displaystyle\frac{-n(n+1)+(n+2)n}{n+2}=\frac{n}{n+2}.$

Plugging in $n=20$ , we find $\displaystyle\sum_{r=1}^{19}\frac{(-1)^{r-1}r}{{20 \choose r}}=\frac{20}{22}=\frac{10}{11},$ so the answer is $\boxed{21}$ .