If $S$ is the sum and $n$ the size, every $\frac{S - x}{n - 1}$ is an integer, so all elements are congruent mod $n - 1$

With $1$ and $2002$ in the set, $n - 1$ divides $2001,$ and $n$ distinct such elements force $(n - 1)^2 + 1 \le 2002$

Let $\mathcal{S}$ have $n$ elements with sum $S.$ The condition says $\frac{S - x}{n - 1}$ is an integer for every $x \in \mathcal{S},$ which means every element is congruent to $S$ modulo $n - 1.$ In particular all elements are congruent to each other, and since $1 \in \mathcal{S},$ every element is $1$ more than a multiple of $n - 1.$

Then $2002 \equiv 1 \pmod{n - 1},$ so $n - 1$ divides $2001 = 3 \cdot 23 \cdot 29.$ Moreover the $n$ distinct elements run from $1$ up to $2002$ in steps that are multiples of $n - 1,$ so $2002 \ge 1 + (n - 1)^2,$ forcing $n - 1 \le 44.$ The largest divisor of $2001$ that is at most $44$ is $29,$ so $n \le 30.$

Thirty is attainable: take the $29$ numbers $1, 30, 59, \ldots, 813$ together with $2002.$ All are $\equiv 1 \pmod{29},$ and the sum of all $30$ is $\equiv 30 \equiv 1 \pmod{29},$ so every deleted mean is an integer. The answer is $30.$