jmc

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 $t\le 1991$, 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 $r=\boxed{990}$.