1. Find the middle term of $(1+x)^{20}$

For $(1+x)^n$, the general term is $T_{r+1} = \binom{n}{r}x^r.$

Since $n=20$ is even, the expansion has $20+1=21$ terms, so there is one middle term, namely the $\left(\frac{21+1}{2}\right)=11^{\text{th}} \text{ term}.$

Thus, for the 11th term, $r=10$. So the coefficient of the middle term is $\binom{20}{10}.$

2. Find the two middle terms of $(1+x)^{19}$

Since $n=19$ is odd, the expansion has $19+1=20$ terms, so there are two middle terms, namely the 10th and 11th terms.

• 10th term: $r=9$, coefficient $\binom{19}{9}$
• 11th term: $r=10$, coefficient $\binom{19}{10}$

Their sum is $\binom{19}{9}+\binom{19}{10}.$

Using Pascal's identity, $\binom{19}{9}+\binom{19}{10}=\binom{20}{10}.$

3. Compute the required ratio

Required ratio: $\frac{\text{coefficient of middle term in }(1+x)^{20}}{\text{sum of coefficients of two middle terms in }(1+x)^{19}}$

So, \frac{\binom{20}{10}}{\binom{19}{9}+\binom{19}{10}}=rac{\binom{20}{10}}{\binom{20}{10}}=1.

4. Final answer

The required integer is $\boxed{1}.$