Go ahead, pick two

If you start with $m$ red balls and $n$ black balls, then the probability that you pick two different balls is given by:

$P(m,n) = \frac{m \times n}{\binom{m+n}{2}}$

This only equals $0.5$ only if $m$ and $n$ are consecutive triangular numbers, i.e. $T_i$ and $T_{i+1}$ where $T_i = 1 + 2 + ... + i$

The only triangular number in the list is $\boxed{210}$

If we let $n$ be the number of balls in the bag and $r$ be the number of red balls we have:

$\dfrac{r}{n} \times \dfrac{n-r}{n-1}+\dfrac{n-r}{n} \times \dfrac{r}{n-1}=0.5$

$\dfrac{2r(n-r)}{n(n-1)}=0.5$

$4r n-4r^2=n^2-n$

$n^2-(4r+1)n+4r^2=0$

Using the quadratic formula we get:

$n=\dfrac{4r+1 \pm \sqrt{(4r+1)^2-16r^2}}{2}$

$n=\dfrac{4r+1 \pm \sqrt{8r+1}}{2}$

Hence it follows that $8r+1$ is a perfect square.Quickly checking the options shows the only suitable value of $r$ is $r=210$ .

$\color{#20A900}{\boxed{\boxed{r=210,n=441 \text{ or } 400}}}$