If we factor $n_k$ into primes, $n_k = p_{1}^{b_{1}}\cdots p_r^{b_{r}}$, then the Chinese Remainder Theorem tells us that $x\equiv a_k\pmod{n_k}$ is equivalent to the system of congruences
$$\begin{align*}
x&\equiv a_k\pmod{p_1^{b_{1}}}\\
x&\equiv a_k\pmod{p_2^{b_{2}}}\\
&\vdots\\
x&\equiv a_k\pmod{p_r^{b_{r}}}
\end{align*}$$
Thus, we can replace the given system of congruences with one in which every modulus is a prime power, $n_i = p_i^{b_i}$.
Note that the assumption that $a_i\equiv a_j\pmod{\gcd(n_i,n_j)}$ "goes through" this replacement (if they were congruend modulo $\gcd(n_i,n_j)$, then they are congruent modulo the gcds of the prime powers as well).
So, we may assume without loss of generality that every modulus is a prime power.
I claim that we can deal with each prime separately, again by the Chinese Remainder Theorem. If we can solve all congruences involving the prime $p_1$ to obtain a solution $x_1$ (which will be determined modulo the highest power of $p_1$ that occurs); and all congruences involving the prime $p_2$ to obtain a solution $x_2$ (which will be determined modulo the highest power of $p_2$ that occurs); and so on until we obtain a solution $x_n$ for all congruences involving the prime $p_n$ (determined modulo the highest power of $p_n$ that occurs), then we can obtain a simultaneous solution by solving the usual Chinese Remainder Theorem system
$$\begin{align*}
x &\equiv x_1 \pmod{p_1^{m_1}}\\
&\vdots\\
x &\equiv x_n\pmod{p_n^{m_n}}
\end{align*}$$
(where $m_i$ is the highest power of $p_i$ that occurs as a modulus).
So we are reduced to solving figuring out whether we can solve the system
$$\begin{align*}
x &\equiv a_1\pmod{p^{b_1}}\\
x &\equiv a_2\pmod{p^{b_2}}\\
&\vdots\\
x & \equiv a_n\pmod{p^{b_n}}
\end{align*}$$
with, without loss of generality, $b_1\leq b_2\leq\cdots\leq b_n$.
When can this be solved? Clearly, this can be solved if and only if $a_i\equiv a_j\pmod{p^{b_{\min(i,j)}}}$: any solution must satisfy this condition, and if this condition is satisfied, then $a_n$ is a solution.
For example: say the original moduli had been $n_1 = 2^3\times 3\times 7^2$, $n_2= 2^2\times 5\times 7$, $n_3=3^2\times 5^3$. First we replace the system with the system of congruences
$$\begin{align*}
x&\equiv a_1 \pmod{2^3}\\
x&\equiv a_2\pmod{2^2}\\
x&\equiv a_1\pmod{3}\\
x&\equiv a_3\pmod{3^2}\\
x&\equiv a_2\pmod{5}\\
x&\equiv a_3\pmod{5^3}\\
x&\equiv a_1\pmod{7^2}\\
x&\equiv a_2\pmod{7}.
\end{align*}$$
Then we separately solve the systems:
$$\begin{align*}
x_1&\equiv a_1 \pmod{2^3} &x_2&\equiv a_1\pmod{3}\\
x_1&\equiv a_2\pmod{2^2}&x_2&\equiv a_3\pmod{3^2}\\
\strut\\
x_3&\equiv a_2\pmod{5}&x_4&\equiv a_1\pmod{7^2}\\
x_3&\equiv a_3\pmod{5^3}&x_4&\equiv a_2\pmod{7}.
\end{align*}$$
Assuming we can solve these, $x_1$ is determined modulo $2^3$, $x_2$ modulo $3^2$, $x_3$ modulo $5^3$, and $x_4$ modulo $7^2$, so we then solve the system
$$\begin{align*}
x &\equiv x_1\pmod{2^3}\\
x &\equiv x_2\pmod{3^2}\\
x&\equiv x_3 \pmod{5^3}\\
x&\equiv x_4\pmod{7^2}
\end{align*}$$
and obtain a solution to the original system.
Hence, if the condition $a_i\equiv a_j\pmod{\gcd(n_i,n_j)}$ holds in the original system, then we obtain a solution for each prime, and from the solution for each prime we obtain a solution to the original system by applying the usual Chinese Remainder Theorem twice.
First Bit - why the solution of the system is a valid answer
If we have a number $y$ such that $y\equiv9\pmod{3}$ and $y\equiv9\pmod{4}$ and $y\equiv9\pmod{23}$ then we know that $y-9\equiv0\pmod{3}$ and $y-9\equiv0\pmod{4}$ and $y-9\equiv0\pmod{23}$. In other words $y-9$ divides by $3$, $4$ and $23$. As these are all coprime then $y-9$ must divide by their product which is $276$.
Second Bit - actually doing the math
$17x\equiv9\pmod{3}$
$17x\equiv0\pmod{3}$
So $x=3y,y\in\mathbb{Z}$
So the second equation becomes
$17x\equiv9\pmod{4}$
$51y\equiv1\pmod{4}$
$3y\equiv1\pmod{4}$
$y\equiv3\pmod{4}$
So $y=4z+3,z\in\mathbb{Z}$
Putting this into the third equation:
$17x\equiv9\pmod{23}$
$51y\equiv9\pmod{23}$
$5y\equiv9\pmod{23}$
$5(4z+3)\equiv9\pmod{23}$
$20z+15\equiv9\pmod{23}$
$20z+6\equiv0\pmod{23}$
This leads easily to $z=2$ which then gives $y=11$ and hence $x=33$.
Checking: $17\times33=561=2\times276+9$
Best Answer
$x=22$ is not a solution, since $22\not\equiv5\pmod6$.