The ship has to weigh more than the water the submerged part of the ship displaces..
But, it would be a good chunk of weight that is there to begin with.

no, that part i am 100% certain of. archimedes' principle. if an object floats the weight of the water it displaces = total object weight.

if an object sinks it'* mass > mass of water displacement.

Yep your right.

i would assume that 4.25 meter is equal to or just greater than the depth of the shallowest of ports functional for ships planned on using water passage.

Gumball, how sure are you of that answer? It doesn't seem logical in a closed system; the displaced water has no place to go so the level raises.

Maybe only one lock is closed at a time? (ie - open system)

when i initially though about it i was in the same mind set as you. from the numbers the bridge is 981 meters long, 34 meters wide and 4.5 deep. that stated you have 132,651 cubic meters of water.

if the ship is 120 meters long by 12 meters wide with it'* hull going 4 meters deep it would only displace 5760 cubic meters of water.

i don't even know the vessels are that big. but in any event the weight of the ship is dispersed through the whole bridge through the water. per square meter on the bridge floor force per ship sized above would only increase ~ 4%

so that stated open loop vs closed loop is negligable.

Now put 10 of those ships on the bridge.
I'm really curious of the physics behind the answer and I'm no mechanical engineer but I'b be willing to bet that only one lock is closed at a time to minimize the load on the bridge. I haven't found anything relevant on google yet either.

I bet that bridge is a "No Wake" zone...