Let's start with cross aldol

Choose the carbanion for the nucleophilic attack in $n$ ways. The aldehyde to be attacked can be chosen in $n-1$ ways as we require cross-aldol products. So, total number of cross-aldol products is given by $\boxed{ n(n-1) }$ For $n=1000$ , the answer comes out to be $\boxed{ 999000 }$

Relevant wiki: Principle of Inclusion and Exclusion - Generalized

Now, since we have $n$ aldehydes, the no. of distinct carbanions are $n$ . Out of these select one carbanion in $\dbinom{n}{1}$ ways and also choose one aldehyde on which it attacks in $\dbinom{n}{1}$ ways. Since both these events must happen by fundamental counting principle of multiplication the total no.of aldol condensation products are

$\dbinom{n}{1}$ $\dbinom{n}{1}$ [Its not $\dbinom{n}{2}$ because the order between the aldehydes matters ] ways. But there are $n$ aldol (non-cross) products. Therefore by Principle of Inclusion and Exclusion the total no. ways is $\dbinom{n}{1}$ $\dbinom{n}{1}$ $-$ $n$ . Substituting $n=1000$ , we get the above answer.