A Sock Problem!!

We have that $\dfrac{m}{n + m}*\dfrac{m - 1}{n + m - 1} + \dfrac{n}{n + m}*\dfrac{n - 1}{n + m - 1} = \dfrac{1}{2}$

$\Longrightarrow 2(m^{2} - m + n^{2} - n) = (n + m)(n + m - 1) = n^{2} + 2mn + m^{2} - n - m$

$\Longrightarrow m^{2} - m + n^{2} - n = 2mn.$

Now let $n = m - k$ for some integer $k \gt 0.$ Then our equation becomes

$m^{2} - m + (m - k)^{2} - (m - k) = 2m(m - k)$

$\Longrightarrow m^{2} - m + m^{2} - 2mk + k^{2} - m + k = 2m^{2} - 2mk$

$\Longrightarrow k^{2} + k - 2m = 0 \Longrightarrow 2m = k(k + 1)$

Now we require that

$m + n = m + (m - k) = 2m - k = k(k + 1) - k = k^{2} \le 101.$

Since $k$ is a positive integer, the maximum value of $k$ is $10$ , giving us

$2m - 10 = 100 \Longrightarrow \boxed{m = 55}$ as the maximum value of $m$ .

Since the probability that they have the same color is $0.5$ , the probability that they have different colors is also $0.5$ . But this probability is simple: $\dfrac{mn}{\binom{m+n}{2}}$ . Thus,

$\dfrac{mn}{\binom{m+n}{2}} = \dfrac{1}{2}$

$2mn = \dfrac{(m+n)(m+n-1)}{2}$

$4mn = (m+n)^2 - (m+n)$

$m+n = (m+n)^2 - 4mn = (m-n)^2$

Now let $m-n = k$ , so $m+n = k^2$ . Thus $m = \frac{k^2+k}{2}$ . For $k \ge 0$ , this function is increasing, so maximizing $m$ is equal to maximizing $k$ .

But $k = \sqrt{m+n}$ is also increasing in $m+n$ , so maximizing $k$ is also maximizing $m+n$ .

Since $m+n \le 101$ and must be a perfect square (so that $k$ is an integer), we have $\max m+n = 100$ , so $k = 10$ and $m = \frac{10^2+10}{2} = \boxed{55}$ .