4. First, for each $i=1,2, \cdots, \frac{p-1}{2}$,
$$\frac{2 i}{p}\binom{p}{2 i}=\frac{(p-1)(p-2) \cdots(p-(2 i-1))}{(2 i-1)!} \equiv \frac{(-1)(-2) \cdots(-(2 i-1))}{(2 i-1)!} \equiv-1 \quad(\bmod p) \text {. }$$

Hence
$$\begin{aligned} \sum_{i=1}^{(p-1) / 2} i^{p-2} & \equiv-\sum_{i=1}^{(p-1) / 2} i^{p-2} \frac{2 i}{p}\binom{p}{2 i} \equiv-\frac{2}{p} \sum_{i=1}^{(p-1) / 2} i^{p-1}\binom{p}{2 i} \\ & \equiv-\frac{2}{p} \sum_{i=1}^{(p-1) / 2}\binom{p}{2 i} \quad(\bmod p) \quad \text { (by Fermat's Little Theorem.) } \end{aligned}$$

The last summation counts the even-sized nonempty subsets of a $p$-element set, of which there are $2^{p-1}-1$.