The messy drawer

Let $r$ be the number of red gloves and $w$ the number of white gloves, with $r \gt w$ and $r + w = N \le 2019$ . The scenario of picking two red gloves in a row or two white gloves in a row translates to the equation

$\dfrac{r}{N} \times \dfrac{r - 1}{N - 1} + \dfrac{w}{N} \times \dfrac{w - 1}{N - 1} = \dfrac{1}{2}$ .

But if the probability of getting the same colour gloves is $1/2$ , then the probability of getting different coloured gloves is also $1/2$ , and thus the scenario of picking a red glove followed by a white glove or a white glove followed by a red glove translates to the equation

$\dfrac{r}{N} \times \dfrac{w}{N - 1} + \dfrac{w}{N} \times \dfrac{r}{N - 1} = \dfrac{1}{2}$ .

Comparing these equations gives us that

$r(r - 1) + w(w - 1) = 2rw \Longrightarrow r^{2} - r + w^{2} - w = 2rw \Longrightarrow r^{2} - 2rw + w^{2} = r + w \Longrightarrow (r - w)^{2} = r + w$ .

Now as $r + w \le 2019$ and since $44^{2} \lt 2019 \lt 45^{2}$ the maximum value for $r - w$ is $44$ , in which case $r + w = 44^{2} = 1936$ , and so

$(r - w) + (r + w) = 44 + 1936 \Longrightarrow 2r = 1980 \Longrightarrow r = 990, w = 990 - 44 = 946$ .

Note that maximizing $r - w$ results in maximizing $r + w$ , which in turn results in maximizing $r$ , and so the desired answer is $\boxed{990}$ .