District Round

Let the total number of votes before Paul voted be $n$, and the number of votes for $A$ before Paul voted be $k$. The percentage of votes for $A$ before Paul voted is $\frac{100k}{n}$, which is a positive integer. After Paul votes for $A$, the number of votes for $A$ becomes $k+1$, and the total number of votes becomes $n+1$. The new percentage is $\frac{100(k+1)}{n+1}$. We are told that this new percentage is exactly one more than the previous percentage:

$$\frac{100(k+1)}{n+1}=\frac{100k}{n}+1$$

Let $p=\frac{100k}{n}$, so $k=\frac{pn}{100}$.

Substitute $k$ into the equation:

$$\frac{100\left(\frac{pn}{100}+1\right)}{n+1}=p+1$$

Multiply both sides by $n+1$:

$$100\left(\frac{pn}{100}+1\right)=(p+1)(n+1)$$

$$pn+100=pn+p+n+1$$

$$100=p+n+1$$

$$p+n=99$$

Recall that $k=\frac{pn}{100}$ must be an integer, and $p$ and $n$ are positive integers with $p+n=99$.

Let $n=99-p$.

Then:

$$k=\frac{p(99-p)}{100}$$

We need $k$ to be an integer, so $p(99-p)$ is divisible by $100$.

Let us try all $p$ from $1$ to $98$ (since $p$ and $n$ are positive integers):

We seek $p$ such that $p(99-p)$ is divisible by $100$.

Let us try $p=19$:

$19(99-19)=19\times 80=1520$, and $1520/100=15.2$ (not integer).

Try $p=80$:

$80(99-80)=80\times 19=1520$, $1520/100=15.2$ (not integer).

Try $p=20$:

$20\times 79=1580$, $1580/100=15.8$ (not integer).

Try $p=25$:

$25\times 74=1850$, $1850/100=18.5$ (not integer).

Try $p=50$:

$50\times 49=2450$, $2450/100=24.5$ (not integer).

Try $p=4$:

$4\times 95=380$, $380/100=3.8$ (not integer).

Try $p=75$:

$75\times 24=1800$, $1800/100=18$ (integer!).

So $p=75$, $n=24$, $k=18$.

Check: Before Paul voted, $n=24$, $k=18$, so percentage is $\frac{100\times 18}{24}=75$.

After Paul votes for $A$, $k=19$, $n=25$, percentage is $\frac{100\times 19}{25}=76$.

So the percentage increases by exactly $1$.

Therefore, Paul's vote was the nineteenth vote for option $A$.