Artificially large odds

This is more of an academic quesiton than a real poker question, because the issue described below is very rare, so in the long-term it hardly matters at all what you do here. Still interested to know how to play this though.

I was just playing poker on pokerstars and they dealt out an anniversary hand (7709000000). This creates special circumstances where the winner of the hand gets $1000 (plus the pot) and everyone else that was dealt in at the table gets $400.

Anyway, when the hand was dealt, an announcement was made to all tables, and it said which table was the lucky winner. I went to that table and it turned out to be a 1/2 NL table.

So, the question is, given these circumstances (with no reads), and assuming you have an ~average stack of $150, if it is folded or limped to you, is your correct action to always push all-in regardless of your hole cards or position?

What if it is raised to you...for the sake of argument, assume someone who acts before you, also with an average stack, pushes all-in. Is it always correct to call with any two in any position?

I actually thought about this exact thing a few days ago. Everyone has already won $400 and the winner of the hand wins an additional $600. So there is basically $600 worth of dead money in the pot. That means you have pot odds to call with pretty much anything and a push to steal that might not be such a bad move.

Lukie recommends folding preflop here.

Don't forget the main pot as an added boon as well

Say youre against 1 guy who show you aces and pushes.

You are calling $150 to win $150 + $600

5:1 odds is good enough to call with any 2.

I think it might become a fold if you hold something reallllly bad like 72o and lots of people push in but I dont have pokerstove on this computer so its hard to check that out. I doubt you can ever have a hand so bad that calling would be more than a tiny mistake and more often than not folding is a huge mistake.

Usually you arent going to be much worse than 2:1. 4:1 is about as bad as you can get (Overpair).