Given that $2\left|\left(1994^{2}-1994\right), 3\right|\left(1994^{3}-\right.$ 1994), we conjecture that $p \mid\left(1994^{p}-1994\right)$.

To prove the above conjecture, we first prove a more general conclusion: If $p$ is a prime, then $p \mid\left(n^{p}-n\right),(n \in N)$.
(1) When $n=1$, it is obvious that $p \mid\left(1^{p}-1\right)$.
(2) Assume that when $n=k$, $p \mid\left(k^{p}-k\right)$.
Then when $n=k+1$,
$$\begin{array}{l} (k+1)^{p}-(k+1) \\ =\sum_{r=0}^{p} C_{p}^{r} \cdot k^{p-r}-(k+1) \\ =\left(k^{p}-k\right)+\sum_{r=1}^{p-1} C_{p}^{r} k^{p-r} . \\ \because r C_{p}^{r}=p C_{p-1}^{r-1}(1 \leqslant r \leqslant p-1), C_{p}^{r}, C_{p-1}^{r-1} \in \end{array}$$
$N$, and $p$ is a prime,
$$\therefore p \mid C_{p}^{r}(1 \leqslant r \leqslant p-1) \text {. }$$

According to the inductive hypothesis, we know
$$p \mid\left[(k+1)^{p}-\{k+1\}\right] \text {, i.e., } n=: k+1 \text {. }$$
is true.
Based on (i) and (2), we can conclude that for $n \in N$, $p \mid\left(n^{p}-n\right)$ always holds.
Thus, $p$ can divide $1994^{p}-1994$.