How to get high current from 9 volt batteries
A 9V battery has approximately a stored energy of:

$$ \require{cancel}
\frac{560\cancel{m}A\cancel{h} \cdot 9V}{1}
\frac{3600s}{\cancel{h}}
\frac{1}{1000\cancel{m}}
\approx 18144VAs
\approx 18 kJ
$$

A joule is a watt-second, or a newton-meter. Under the most ideal conditions, with perfectly efficient machines everywhere, there is enough stored energy in a 9V battery to apply your specified 600N force over a distance of:

$$ \frac{18\cancel{k}\cancel{J}}{1}
\frac{\cancel{N}m}{\cancel{J}}
\frac{1}{600\cancel{N}}
\frac{1000}{\cancel{k}}
= 30m
$$

Your proposed solenoid, requiring maybe \$25A\$ at \$9V\$, consumes electrical power at the rate of:

$$ 25A \cdot 9V = 225W $$

Applying your specified \$600N\$ force, and given that power, we can solve for the velocity your solenoid, were it 100% efficient, could provide:

$$ \frac{225\cancel{W}}{1}
\frac{\cancel{J}}{\cancel{W}s}
\frac{\cancel{N}m}{\cancel{J}}
\frac{1}{600\cancel{N}}
= 0.375m/s
$$

So you see, even if we can extract all the stored energy of the 9V battery with 100% efficiency, there isn't a whole ton of it. Knowing that your ideal solenoid is moving at \$0.375m/s\$, and that the battery has enough energy for it to move \$30m\$, the runtime is:

$$ \frac{30\cancel{m}}{1}
\frac{s}{0.375\cancel{m}}
= 80s $$

Or we could calculate it from the battery energy and solenoid power:

$$ \frac{18000\cancel{W}s}{1}
\frac{1}{225\cancel{W}}
= 80s
$$

But maybe it's enough. The question is how to do it efficiently. Electrical power in a resistance is given by:

$$ P = I^2 R$$

The internal resistance of a 9V battery is maybe \$1.5\Omega\$, when fresh. It goes up as the battery drains. Your solenoid is probably at least another \$1\Omega\$. So at \$25A\$, your resistive losses alone would be:

$$ (25A)^2 (1.5\Omega + 1\Omega) = 1562.5W$$

Compare this to the power used by the ideal solenoid considered above (\$225W\$) and you can see this is an absurdly inefficient system. Just dealing with the heat from these losses will be a challenge. Of course, you can't actually get this out of a 9V battery, because the voltage lost over its internal resistance at \$25A\$ is:

$$ 25A \cdot 1.5\Omega = 37.5V $$

...which is more than the 9V supplied by the battery.

Besides the battery, or the solenoid, transferring \$225W\$ of electrical power is a problem itself. Because power is the product of voltage and current (\$P=IE\$), to move a lot of power you can have high current, or high voltage. But, even wires have resistance, and since power lost to this resistance is proportional to the square of current, it's more practical to move high amounts of electrical power at high voltage than it is at high current. This is why the electric utility transmits power over long distances at very high voltage.

So, if you want to move \$225W\$ at \$9V\$, you must keep the resistance very low, to avoid resistive losses from being very high. That means fat wire (including the wire in your solenoid, which accounts for most of the wire in the circuit) and batteries with low internal resistance. You can also trade current for voltage, or voltage for current, in the design of your solenoid, as supercat's answer describes.