Is this simple problem NP-complete?
First, we prove that the decision version of the original problem is NP. Since an instance of an assignment can be verified within polynomial time, it's obviously NP.

Now, we try to prove the NP-hardness of the problem.
Well, I believe this problem can be reduced from the bin-packing problem.

The bin-packing problem says the following:
There are $n$ items each with a weight $w_i$ and $m$ bins. We want to put the items to bins. Each bin has a maximum weight limit $\omega$. The problem is to find a item assignment scheme so as to minimize the number of bins needed. Mathematically:
$$\textrm{minimize}~~~ z = \sum_{j=1}^m y_j$$
Subject to:
$$\sum_{i=1}^n w_i x_{i,j} \leq \omega y_j, \forall j \in \{1,2,\cdots,m\}$$
$$\sum_{j=1}^m x_{i,j} = 1, \forall i \in \{1,2,\cdots,n\}$$
$$y_j \in \{0,1\}, \forall j$$
$$x_{i,j} \in \{0,1\}, \forall i, \forall j$$
The problem is NP-hard, and its NP-hardness has already been proved. The decision version of this problem is NP-complete. 

Note that in the candy allocation problem, we tend to maximize the number of kids that are completely full. Since the number of kids are the same, the problem is equivalent to minimizing the number of kids that are not full, while ensuring every candy is allocated. The problem is formalized as follow:
$$\textrm{minimize}~~~ \sum_{j=1}^m y_j$$
Subject to:
$$\sum_{i=1}^n x_{i,j}w_i \leq \omega y_j, \forall j \in \{1,\cdots,m\}$$
$$\sum_{j=1}^m x_{i,j}=1 ,\forall i \in \{1,\cdots,n\}$$
$$x_{i,j} \in \{0,1\}, \forall i, \forall j$$
$$y_j \in \{0,1\}, \forall j$$
where $y_j$ denotes if kid $j$ is hungry ("1" is hungry, "0" is full), and $x_{i,j}$ denotes if candy $i$ is allocated to kid $j$.

One can easily see that the problem is equivalently to the bin packing problem. Thus, candy allocation is NP-hard.

Since candy allocation is NP and NP-hard, it's NPC. 

The definition of bin packing problem can also be found in https://en.wikipedia.org/wiki/Bin_packing_problem.