Yes; and this problem is just about making notations before you decide to try anything else. Write $c_j$ for the number of purses in pocket $j$, and $s_j$ for the total sum in pocket $j$. We are given $\left\{ \begin{array}{l} c_1 + \cdots + c_M = N \\ s_1 + \cdots + s_M = 2003 \end{array} \right.$. Now, if any purse had at least $M$ dollars in it, then for all $j$, $N > s_j \ge c_jM$. Summing up for all $j \in \{1, \ldots, M\}$, $NM > (c_1 + \cdots + c_M)M = NM$, contradiction. $2003$ was prime, but just misleading here.

The answer is yes. For the sake of contradiction, suppose that all purses have at least $M$ dollars. Then $2003\ge MN$ because there are $N$ purses. On the other hand, there exists a pocket with at least $\frac{2003}{M}$ dollars, so $N>\frac{2003}{M}\implies MN>2003$, a contradiction.