Only numbers!

$\begin{pmatrix} 2000 \\ 1000 \end{pmatrix} = \dfrac{2000 \times 1999 \times 1998 \times ... \times 1001}{1 \times 2 \times 3 \times ... \times 1000}$

All the divisors including the primes are cancelled out. The 3-digit primes in the nominator not cancelled out are those $< 2000 \div 3$ and the largest of which is $\boxed{661}$ .

If $p$ is a prime such that $666\quad <\quad p\quad <\quad 1000$ , then the only numbers $\le 2000$ , which are divisible by
$p$ are $p$ and $2p$ . Therefore the highest power of $p$ which divides $2000!$ is ${ p }^{ 2 }$ . Obviously, the highest power of $p$ dividing $1000!$ is $p$ itself.

Since ${2000 \choose 1000}$ = $\frac { 2000! }{ { (1000!) }^{ 2 } }$ , $p$ does not divide
${2000 \choose 1000}$ .

So let $p$ be the largest prime less than $666$ , viz., $661$ .

It is obvious that the highest powers of $p$ which divide $2000!$ and $({ 1000!) }^{ 2 }$ are ${ p }^{ 3 }$ and ${ p }^{ 2 }$ respectively; therefore $p$ divides ${2000 \choose 1000}$ .

So. answer : $\boxed{661}$ .