4. In the tetrahedron $P-ABC$,
$$
\begin{array}{l}
\angle APB=\angle BPC=\angle CPA=90^{\circ}, \\
PA=4, PB=PC=3 .
\end{array}
$$

Then the minimum value of the sum of the squares of the distances from any point on the base $ABC$ to the three side faces is
4. $\frac{144}{41}$.

Since the sum of the squares of the distances from the inner point $M$ on the base to the three side faces is $M P^{2}$, when $M P \perp$ plane $A B C$, $M P$ is minimized, let the minimum value be $x$.
$$
\begin{array}{l}
\text { Given } S_{\triangle A B C}=\frac{3 \sqrt{41}}{2}, V_{\text {tetrahedron } B-P A C}=V_{\text {tetrahedron } P-A B C} \\
\Rightarrow x=\frac{12}{\sqrt{41}} \Rightarrow x^{2}=\frac{144}{41} .
\end{array}
$$