jmc

Suppose $\lfloor \frac{2002}{n}\rfloor =k.$ 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 $\lfloor \frac{2002}{n}\rfloor =k$ 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 $$\begin{array}{rl}\lfloor \frac{2002}{44}\rfloor & =45,\\ \lfloor \frac{2002}{43}\rfloor & =46,\\ \lfloor \frac{2002}{42}\rfloor & =47,\\ \lfloor \frac{2002}{41}\rfloor & =48.\end{array}$$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 $\boxed{49}.$