A drawer contains a mixture of red socks and blue socks, at most $1991$ in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data?
Solution 1
Let $r$ and $b$ denote the number of red and blue socks, respectively. Also, let $t=r+b$. The probability $P$ that when two socks are drawn randomly, without replacement, both are red or both are blue is given by
\[\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.\]
Solving the resulting quadratic equation $r^{2}-rt+t(t-1)/4=0$, for $r$ in terms of $t$, one obtains that 
\[r=\frac{t\pm\sqrt{t}}{2}\, .\]
Now, since $r$ and $t$ are positive integers, it must be the case that $t=n^{2}$, with $n\in\mathbb{N}$. Hence, $r=n(n\pm 1)/2$ would correspond to the general solution. For the present case $t\leq 1991$, and so one easily finds that $n=44$ is the largest possible integer satisfying the problem conditions. 
In summary, the solution is that the maximum number of red socks is $r=\boxed{990}$.

Solution 2
Let $r$ and $b$ denote the number of red and blue socks such that $r+b\le1991$. Then by complementary counting, the number of ways to get a red and a blue sock must be equal to $1-\frac12=\frac12=\frac{2rb}{(r+b)(r+b-1)}\implies4rb=(r+b)(r+b-1)$
$=(r+b)^2-(r+b)\implies r^2+2rb+b^2-r-b=4rb\implies r^2-2rb+b^2$
$=(r-b)^2=r+b$, so $r+b$ must be a perfect square $k^2$. Clearly, $r=\frac{k^2+k}2$, so the larger $k$, the larger $r$: $k^2=44^2$ is the largest perfect square below $1991$, and our answer is $\frac{44^2+44}2=\frac12\cdot44(44+1)=22\cdot45=11\cdot90=\boxed{990}$.


Solution 3
Let $r$ and $b$ denote the number of red and blue socks, respectively. In addition, let $t = r + b$, the total number of socks in the drawer.
From the problem, it is clear that $\frac{r(r-1)}{t(t-1)} + \frac{b(b-1)}{t(t-1)} = \frac{1}{2}$
Expanding, we get $\frac{r^2 + b^2 - r - b}{t^2 - t} = \frac{1}{2}$
Substituting $t$ for $r + b$ and cross multiplying, we get $2r^2 + 2b^2 - 2r - 2b = r^2 + 2br + b^2 - r - b$
Combining terms, we get $b^2 - 2br + r^2 - b - r = 0$
To make this expression factorable, we add $2r$ to both sides, resulting in $(b - r)^2 - 1(b-r) = (b - r - 1)(b - r) = 2r$
From this equation, we can test values for the expression $(b - r - 1)(b - r)$, which is the multiplication of two consecutive integers, until we find the highest value of $b$ or $r$ such that $b + r \leq 1991$. 
By testing $(b - r - 1) = 43$ and $(b - r) = 44$, we get that $r = 43(22) = 946$ and $b = 990$. Testing values one integer higher, we get that $r = 990$ and $b = 1035$. Since $990 + 1035 = 2025$ is greater than $1991$, we conclude that $(946, 990)$ is our answer. 
Since it doesn't matter whether the number of blue or red socks is $990$, we take the lower value for $r$, thus the maximum number of red socks is $r=\boxed{990}$.

Solution 4
As above, let $r$, $b$, and $t$ denote the number of red socks, the number of blue socks, and the total number of socks, respectively. We see that $\frac{r(r-1)}{t(t-1)}+\frac{b(b-1)}{t(t-1)}=\frac{1}{2}$, so $r^2+b^2-r-b=\frac{t(t-1)}{2}=r^2+b^2-t=\frac{t^2}{2}-\frac{t}{2}$. 
Seeing that we can rewrite $r^2+b^2$ as $(r+b)^2-2rb$, and remembering that $r+b=t$, we have $\frac{t^2}{2}-\frac{t}{2}=t^2-2rb-t$, so $2rb=\frac{t^2}{2}-\frac{t}{2}$, which equals $r^2+b^2-t$.
We now have $r^2+b^2-t=2rb$, so $r^2-2rb+b^2=t$ and $r-b=\pm\sqrt{t}$. Adding this to $r+b=t$, we have $2r=t\pm\sqrt{t}$. To maximize $r$, we must use the positive square root and maximize $t$. The largest possible value of $t$ is the largest perfect square less than 1991, which is $1936=44^2$. Therefore, $r=\frac{t+\sqrt{t}}{2}=\frac{1936+44}{2}=\boxed{990}$.
Solution by Zeroman

Solution 5
Let $r$ be the number of socks that are red, and $t$ be the total number of socks. We get:
$2(r(r-1)+(t-r)(t-r-1))=t(t-1)$
Expanding the left hand side and the right hand side, we get:
$4r^2-4rt+2t^2-2t = t^2-t$
And, moving terms, we will get that:
$4r^2-4rt+t^2 = t$
We notice that the left side is a perfect square. 
$(2r-t)^2 = t$
Thus $t$ is a perfect square. And, the higher $t$ is, the higher $r$ will be. So, we should set $t = 44^2 = 1936$
And, we see, $2r-1936 = \pm44$ We will use the positive root, to get that $2r-1936 = 44$, $2r = 1980$, and $r = \boxed{990}$.
- AlexLikeMath