Find the least positive integer $k$ for which the equation $\left\lfloor\frac{2002}{n}\right\rfloor=k$ has no integer solutions for $n$. (The notation $\lfloor x\rfloor$ means the greatest integer less than or equal to $x$.)
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 $\boxed{49}.$