Solution:
Since $\{c_{n}\}$ is a strictly increasing sequence of positive integers, it is clear that $c_{n} \geq n, n \in \mathbb{N}$. Hence, $c_{a_{n}} \geq a_{n}, n \in \mathbb{N}$. However, the given sequences do not contain equal terms, so $c_{a_{n}}>a_{n}$ and $b_{n}=c_{a_{n}}-1>a_{n}, n \in \mathbb{N}$. Similarly, from (ii) and (iii), $a_{n+1}>b_{n}+1=c_{a_{n}}, n \in \mathbb{N}$. It is also easy to see that $b_{n}<c_{a_{n}}<b_{n+1}$. Let us for any $n \in \mathbb{N}$ count the number of terms in all three sequences that are less or equal to $c_{a_{n}}$. There are $n$ such terms in the first sequence (that is, $a_{1}, a_{2}, \ldots, a_{n}$ ), $n$ such terms in the second sequence $\left(b_{1}, b_{2}, \ldots, b_{n}\right)$ and $a_{n}$ such terms in the third sequence $\left(c_{1}, c_{2}, \ldots, c_{a_{n}}\right)$. It is $2 n+a_{n}$ terms in total. By (i) any positive integer less or equal $c_{a_{n}}$ must appear among these terms exactly once; thus, the total number of these terms equals
$$2 n+a_{n}=c_{a_{n}}$$
Now we take $a_{n}$ instead of $n$ in (iv):
$$\begin{aligned} & c_{a_{n}+1} c_{a_{n}}-\left(a_{n}+1\right) c_{a_{n}+1}-a_{n} c_{a_{n}}=c_{a_{n}+1}\left(a_{n}+2 n\right)-\left(a_{n}+1\right) c_{a_{n}+1}-a_{n}\left(a_{n}+2 n\right)= \\ & =c_{a_{n}+1}(2 n-1)-a_{n}^{2}-2 a_{n} n \equiv c_{a_{n}+1}-a_{n}-2 n \equiv c_{a_{n}+1}-c_{a_{n}} \equiv 0 \quad(\bmod 2) \end{aligned}$$
This means that $c_{a_{n}+1} \neq c_{a_{n}}+1$ and the number $c_{a_{n}}+1$ has to belong to either of the first two sequences. The inequalities $a_{n}<b_{n}<c_{a_{n}}<a_{n+1}<b_{n+1}$ imply that $c_{a_{n}}+1=a_{n+1}$, $n \in \mathbb{N}$, and, by (1),
$$a_{n+1}=a_{n}+2 n+1, \quad n \in \mathbb{N}$$
Next we prove that $a_{1}=1$. Indeed, number 1 has to belong to one of the given sequences, and if $a_{1}>1$ then $c_{1}=1$ or $b_{1}=1$. The latter case is impossible because $b_{1}>a_{1}$. Then we must have $c_{1}=1$, and either $c_{2}=2$, or $a_{1}=2$ and $c_{a_{1}}=c_{2}=a_{1}+2=4$. In both cases we obtain a contradiction by setting $n=1$ in (iv). This proves that $a_{1}=1$, and, together with (2), defines a unique sequence $\{a_{n}\}$ :
$a_{n}=a_{n-1}+(2 n-1)=a_{n-2}+(2 n-3)+(2 n-1)=\cdots=a_{1}+3+5+\ldots+(2 n-1)=n^{2}, \quad n \in \mathbb{N}$.
Hence,
$$\begin{aligned} a_{2010} & =2010^{2} \\ b_{2010} & =c_{a_{2010}}-1=a_{2010}+2 \cdot 2010-1=2011^{2}-2 \\ c_{1936} & =c_{44^{2}}=c_{a_{44}}=a_{44}+2 \cdot 44=44^{2}+88=2024 \\ a_{45} & =45^{2}=2025 \end{aligned}$$
and all the integers between $a_{45}$ and $b_{45}=c_{a_{45}}-1=a_{45}+2 \cdot 45-1=a_{45}+89$ belong to the sequence $\{c_{n}\}$. Hence, these integers have the form
$$c_{1936+k}=a_{45}+k, \quad k=1,2, \ldots, 88$$
and $c_{2010}=c_{1936+74}=a_{45}+74=2099$.
Answer. $a_{2010}=2010^{2}, b_{2010}=2011^{2}-2, c_{2010}=2099$.

Solution:
Denote by $(*)$ the trivial fact $a_{n}<c_{a_{n}}$ derived at the beginning of the first solution. One can easily fill the sequences inductively. In fact, like in the first solution, we have $a_{1}=1$. Now we will find the place for number 2. If $a_{2}=2$, then by (iii) $2=a_{2}>b_{1}$, which is impossible. If $c_{1}=2$, then by (ii) we have $2=c_{1}=c_{a_{1}}=b_{1}+1$, hence $b_{1}=1$ which is also impossible. So the only way is to put $b_{2}=2$. Then by (ii) $c_{1}=c_{a_{1}}=b_{1}+1=3$.