Took a while, but here's my attempt!
1) Starts with 1000L of milk.
Sells 500L to 250 customers, as each buy 2L.
2) Milkman has 500L of pure milk remaining, which he tops up with 500L of water.
The ratio of the Milk:Water mixture is now 1:1 (within the realm of what he can sell) and the total volume is back up to 1000L.
He sells 500L to 250 customers.
3) Milkman has 500L total of diluted milk remaining, consisting of 250L of milk and 250L of water.
He tops up with 500L of water to a total of 1000L, so there's 250L of milk and 750L of water, for a ratio of 1:3 of Milk:Water (still within the realm of what he can sell).
He sells 500L to 250 customers.
4) Milkman has 500L of diluted milk remaining, consisting of 125L of milk and 375L of water.
He tops up with 500L of water to a total of 1000L, so there's 125L of milk and 875L of water, for a ratio of 1:7 of Milk:Water (just within the realm of what he can sell, the 1:7 limit).
Since he won't be able to dilute it any further, he sells all remaining 1000L of diluted milk, to 500 customers.
Now comes the trap to avoid - add all the customers in bold above and you get 1250... which is not one of the answers.
The question is asking how many customers he cheated, not how many he sold to. He didn't actually cheat the first 250 as they bought pure milk, so the correct answer IMO is E, 1000 cheated customers.
Let me know if you can see any shortcuts, as this took me way too long to figure out (and yes, I fell into the trap so didn't even get it right the first time!)