a) Suppose that there are $m$ students labeled by $1,2,3,\ldots,m$ and $11$ types of candy denoted by $a_1,a_2,a_3,\ldots,a_{11}$. Let $X=\{a_1,a_2,a_3,\ldots,a_{11}\}$ and $A_1,A_2,\ldots,A_m$ be the set of candy types that the students $1,2,3,\ldots,m$ received, respectively. We have $A_1,A_2,\ldots,A_m \subset X$.

For each $i=1,2,3,\ldots,11$, let $d(a_i)$ be the number of $a_i$ candies in the total of $2013$ candies. It is clear that
$$d(a_1)+d(a_2)+d(a_3)+\ldots+d(a_{11})=2013.$$
We also have $M = \sum_{1 \le i < j \le m} |A_i \cap A_j|$. For each $i=1,2,3,\ldots,11$ then the number of subsets containing $a_i$ is $d(a_i)$, so the number of two subsets having a common element $a_i$ is $\binom{d(a_i)}{2}$. Therefore, we have
$$M = \sum_{1 \le i < j \le m} |A_i \cap A_j| = \sum_{i=1}^{11} \binom{d(a_i)}{2} = \frac{1}{2} \sum_{i=1}^{11} \left( (d(a_i))^2 - d(a_i) \right).$$
By the Cauchy-Schwarz inequality, we have $\sum_{i=1}^{11} 1^2 \cdot \sum_{i=1}^{11} (d(a_i))^2 \ge \left(\sum_{i=1}^{11} d(a_i)\right)^2 = 2013^2$,
so
$$M \ge \frac{1}{2} \left( \frac{2013^2}{11} - 2013 \right).$$
The equality occurs if and only if $d(a_1) = d(a_2) = d(a_3) = \dots = d(a_{11}) = \frac{2013}{11} = 183$. In this case, there always exists a certain number of students that satisfy this condition.

b.
Similarly as in the previous case, let $b_1, b_2, b_3, \dots, b_9$ be the types of candies and $d(b_i), i = 1, 9$ be the number of candies of each type. We have
$$d(b_1) + d(b_2) + d(b_3) + \dots + d(b_9) = 2013.$$
We need to find the minimum value of $\sum_{i=1}^{9} (d(b_i))^2$.
Notice that if there exists $d(b_i) - d(b_j) \ge 2$ for some $1 \le i, j \le 9$ then we can decrease $d(b_i)$ by $1$ and increase $d(b_j)$ by $1$ then
$$(d(b_i))^2 + (d(b_j))^2 - (d(b_i) - 1)^2 - (d(b_j) + 1)^2 = 2(d(b_i) - d(b_j)) > 0.$$
Therefore, to obtain the minimum value, we need $d(b_i) - d(b_j) \le 1$ for all $i, j \in \{1, 2, 3, \dots, 9\}$.
Without loss of generality, we assume that $d(a_1) \le d(a_2) \le d(a_3) \le \dots \le d(a_9)$. From the above argument, $d(a_i)$ can be either $k, k+1$ for some positive integer $k$. Suppose that there are $t$ numbers $k$ and $9-t$ numbers $k+1$. We need to find the minimum value of
$$M = \frac{1}{2} (tk^2 + (9-t)(k+1)^2 - 2013) \text{ with } tk + (9-t)(k+1) = 2013 \Leftrightarrow t = 9k - 2004.$$
Substitute into $M$, we have
$$M = \frac{1}{2}((9k - 2004)k^2 + (2013 - 9k)(k+1)^2 - 2013).$$
Since $0 \le t \le 9$, we have $0 \le 9k - 2004 \le 9 \Rightarrow \frac{2004}{9} \le k \le \frac{2013}{9} \Rightarrow k = 223$.
With $k=223$ then $M = \frac{1}{2}(3 \cdot 223^2 + 6 \cdot 224^2 - 2013)$.
In this case, the minimum value is $M = \frac{1}{2}(3 \cdot 223^2 + 6 \cdot 224^2 - 2013)$ which can be obtained when we have $3$ numbers $223$ and $6$ numbers $224$.