Suppose on the contrary that no club contains at least $11$ male and $11$ female members. We shall count the number of triples $(M, F, C)$ where $M$ is a male student, $F$ is a female student and $C$ is a club joined by these two students.

Firstly, for each of the $2007$ choices of $M$ and each of the $2007$ choices of $F$, the condition shows that there is at least one club $C$ for which $(M, F, C)$ should be counted. This gives at least $2007^2$ triples.

Secondly, we count according to the clubs. We label the clubs as $C_1, C_2, \dots, C_k$. For each $i = 1, 2, \dots, k$, let $m_i$ and $f_i$ be the numbers of male students and female students in the club $C_i$ respectively. Then the number of triples $(M, F, C)$ is
$$\sum_{i=1}^{k} m_i f_i .$$
From the assumption, we can divide these clubs into two groups. The first group consists of those clubs with at most $10$ male members (say $C_1, C_2, \dots, C_\ell$). The second group consists of those clubs with at least $11$ male members, and hence at most $10$ female members (say $C_{\ell+1}, C_{\ell+2}, \dots, C_k$). Then
$$\begin{aligned} \sum_{i=1}^{k} m_i f_i &= \sum_{i=1}^{\ell} m_i f_i + \sum_{i=\ell+1}^{k} m_i f_i \\ &\le 10 \sum_{i=1}^{\ell} f_i + 10 \sum_{i=\ell+1}^{k} m_i \\ &\le 10 \sum_{i=1}^{k} f_i + 10 \sum_{i=1}^{k} m_i. \end{aligned}$$
Note that the sum over $f_i$ counts the total number of female members in all clubs. By another simple double counting, this is equal to the total number of clubs joined by all the female students. Therefore, we have
$$\sum_{i=1}^{k} f_i \le 100 \times 2007.$$
The same upper bound holds for the sum over $m_i$. Combining all these, we obtain
$$2007^2 \le 10 \times 100 \times 2007 + 10 \times 100 \times 2007 = 2000 \times 2007.$$
This is a contradiction. Hence, there must be a club with at least $11$ male and $11$ female members.