$\left\lfloor \frac{2002}{n} \right\rfloor = k$ has a solution exactly when some integer $n$ lies in $\left(\frac{2002}{k+1}, \frac{2002}{k}\right]$

That interval has length $\frac{2002}{k(k+1)},$ so it always contains an integer when $k(k+1) \le 2002.$ Then compute $\lfloor \frac{2002}{n} \rfloor$ for $n = 44,$ $43,$ $42,$ $41,$ $40.$

The value $k$ is attained exactly when some integer $n$ satisfies $k \le \frac{2002}{n} \lt k + 1,$ that is, when the interval $\left(\frac{2002}{k+1}, \frac{2002}{k}\right]$ contains an integer. Its length is $\frac{2002}{k(k+1)},$ which is at least $1$ whenever $k(k+1) \le 2002$ — so every $k \le 44$ is attained.

For larger $k,$ check directly: $n = 44,$ $43,$ $42,$ $41,$ $40$ give $\left\lfloor \frac{2002}{n} \right\rfloor = 45,$ $46,$ $47,$ $48,$ $50.$ Since $\frac{2002}{41} \approx 48.8$ and $\frac{2002}{40} \gt 50,$ the value $49$ is never attained, so the least such $k$ is $49.$