Prelim: The definition is confusing because it is not assuming $k = n$. You will be able to prove $k = n$ later but in mathematics we don't include anything in a definition that we can prove later.
The definition of a complete residue system is a collection of integers $\{a_j\}$ so that for any integer, that integer will be congruent (have the same remainder) with exactly one of those integers.
In Other words, and probably a much more straightforward definiton, For every possible remainder, there will be be exactly one integer with that remainder.
For instance if $n = 7$, the easiest and most obvious complete residue system would be simply $\{0,1,2,3,4,5,6\}$. Every integer will be have remainder $0,1,2,3,4,5,6$ and those are precisely the numbers in there.
Another complete system could be $\{63, 8, 15, -4, 32, 75, 146\}$. If an integer $b$ has remainder $0$ it is congruent to $63$. $63$ represents all the integers with remainder $0$. ANd if $b$ has remainder $8$ then $b$ is congruent to $8$. $8$ represents all the integers with remainder $1$.... And so on.
Every remainder is represented exactly once.
And that is what a complete residue system means. A residue is a representation of one class of remainders (all the integers with remainder $4$ for example are represented by $32\equiv 4 \pmod 7$, for example). And a complete residue system means every residue is represented.
And as there $n$ possible remainders there will be $n$ elements in the system so if the system is $\{a_1, ...., a_k\}$ then $k = n$. (If it were me, I wouldn't even bring up the idea this could be in doubt. It just confuses things the first time you see the definition.)
.....
Okay. To do a completer residue system $\pmod 4$ you need to find a $\{a_1, a_2, ..... , a_k\}$ were for every integer $-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,....$ is congruent to exact one of them.
So we need and $a_1\equiv -6\pmod 4$. Well $-6\equiv 2 \pmod 4$ so let's let $a_1 = 2$. And we need something $\equiv -5 \pmod 4$. We $-5\equiv 3 \pmod 4$ and $15 \equiv 3 \pmod 4$ and $15 \equiv -3 \pmod 4$ so lets use $a_2 = 15$.
And we need something $\equiv - 4 \pmod 4$. Well $-4 \equiv 0 \equiv 28\pmod 4$ so let's use $a_3 = 28$. And we need something $\equiv -3\pmod 4$ but $-3\equiv 1 \equiv 48321 \pmod 4$. SO lets let $a_4 = 48321$.
And we need something $\equiv -2\pmod 4$. But $-2 \equiv 2 = a_1$ so we already have it. In fact it looks like we have one of each.
So $\{2, 15,28, 48321\}$ seems to be complete.
If $b$ is an integer we have either $b = 4k$ or $4k + 1$ or $4k + 2$ or $4k + 3$.
If $b = 4k$ then $b \equiv 28\pmod 4$. ANd if $4k + 1$ then $b\equiv 48321$. And if $b = 4k + 2$ then $b \equiv 2\pmod 4$ and if $b= 4k + 3$ then $b\equiv 15\pmod 4$.
So ....
the definition is:
$\{2, 15,28, 48321\}$ is called a canonical complete residue system modulo n if every integer is congruent modulo $4$ to exactly one element of the set
Well, that is true so $\{2, 15,28, 48321\}$ is a complete residue system.
That's all.
Best Answer
Let's assume $p<q$. You can only conclude that the the $a_i$'s form a complete system of residues modulo $p$. Let's note that $\displaystyle \sum_{k=0}^{pq-1} k^t\equiv 0 \pmod{p}$ if and only if $p-1\nmid t$ and $\displaystyle \sum_{k=0}^{pq-1} k^t\equiv -q\pmod{p}$ for $p-1|t$. Same goes for switching $p$ and $q$.
Inductively one can show that $\displaystyle \sum_{i=1}^{p} a_i^{n}\equiv 0 \pmod{p}$ for $n\leq p-2$;this formalizes mathworkers post . More precisely if we denote with $\displaystyle A_n=\sum_{i=1}^p a_i^n$ and $\displaystyle B_n=\sum_{i=1}^q b_i^n$ we have $$\sum_{i,j} (a_i+b_j)^n=\sum_{k=1}^{n-1} A_k B_{n-k} \dbinom{n}{k}+qA_n+pB_n$$
Now for $n=p-1$ we have $\displaystyle \sum_{k=0}^{pq-1} k^t\equiv -q \pmod{pq}$ and thus $A_{p-1}\equiv -1\pmod{p}$. Now Fermat's little theorem gives $a_i^{p-1} \equiv \{0,1\} \pmod{p}$ and thus we must have one number divisible by $p$ and $p-1$ coprime with $p$ amongst the $a_i$. Wlog $p|a_1$.
Now considering the polynomial $\displaystyle f(X)=\prod_{i=2}^{p} (X-a_i)$ in $\mathbb{F}_p[X]$ we can show that all the coefficients are zero; these are symmetric sums in the $a_i$ and using Newton's relations between power sums and symmetric sums. Thus $f(X)=X^{p-1}-a$ and since $a_{i}^{p-1}\equiv 1 \pmod{p}$ we have $a=1$. Since $\displaystyle X^{p-1}-1=\prod_{j=1}^{p-1} (X-j) $ in $\mathbb{F}_p[X]$ we have shown that $\{a_i\}_{i=1}^p$ form a complete system of residues modulo $p$.
Note that the same strategy guarantees that $q|B_n$ for $n<q-1$ using the same recurrence relation. By the same argument as above one gets that $\{b_i\}$ form a complete system of residues mod $q$ also.