Solution:

Condition (1) requires the seats $a, 2a, \ldots, na$ to be pairwise distinct modulo $n,$ which happens if and only if $\gcd(a, n) = 1.$ For condition (2), the two mathematicians from seats $i$ and $j$ have gap counts before the break determined by $\pm(i - j) \bmod n$ and after the break by $\pm a(i - j) \bmod n,$ so the requirement is $a(i - j) \not\equiv \pm(i - j) \pmod{n}$ for all $i \ne j.$ Equivalently, $(a - 1)(i - j) \not\equiv 0$ and $(a + 1)(i - j) \not\equiv 0 \pmod{n}$ for every nonzero residue $i - j,$ which holds exactly when $a - 1$ and $a + 1$ are also relatively prime to $n.$

So $n$ is possible if and only if some $a$ satisfies $\gcd\big((a - 1)\,a\,(a + 1),\, n\big) = 1.$ Any three consecutive integers include a multiple of $2$ and a multiple of $3,$ so no $a$ works when $\gcd(n, 6) \gt 1.$ Conversely, if $\gcd(n, 6) = 1,$ then $a = 3$ works, since $2 \cdot 3 \cdot 4 = 24$ has only the prime factors $2$ and $3.$

The valid $n$ with $1 \lt n \lt 1000$ are those congruent to $\pm 1 \pmod 6,$ namely $6k \pm 1$ for $1 \le k \le 166,$ and there are $2 \cdot 166 = 332$ of them.