Suppose $ \left\lfloor \frac{2002}{n} \right\rfloor = k $. Then \[k \le \frac{2002}{n} < k + 1.\]This is equivalent to \[\frac{1}{k + 1} < \frac{n}{2002} \le \frac{1}{k},\]or \[\frac{2002}{k + 1} < n \le \frac{2002}{k}.\]Thus, the equation $ \left\lfloor \frac{2002}{n} \right\rfloor = k $ has no solutions precisely when there is no integer in the interval \[\left( \frac{2002}{k + 1}, \frac{2002}{k} \right].\]The length of the interval is \[\frac{2002}{k} - \frac{2002}{k + 1} = \frac{2002}{k(k + 1)}.\]For $ 1 \le k \le 44, $ $ k(k + 1) < 1980, $ so $ \frac{2002}{k(k + 1)} > 1 $. This means the length of the interval is greater than 1, so it must contain an integer. We have that \begin{align*} \left\lfloor \frac{2002}{44} \right\rfloor &= 45, \\ \left\lfloor \frac{2002}{43} \right\rfloor &= 46, \\ \left\lfloor \frac{2002}{42} \right\rfloor &= 47, \\ \left\lfloor \frac{2002}{41} \right\rfloor &= 48.\end{align*}For $ k = 49, $ the interval is \[\left( \frac{2002}{50}, \frac{2002}{49} \right].\]Since $ 40 < \frac{2002}{50} < \frac{2002}{49} < 41, $ this interval does not contain an integer. Thus, the smallest such $ k $ is $ \boxed{49} $.