4. Let $x$, $y$, $z$ be the lengths of the line segments connecting the midpoints of the opposite edges of the tetrahedron $P-ABC$. If the sum of the squares of the six edges of the tetrahedron is 300, then $x^{2}+y^{2}+z^{2}=$ $\qquad$ .
4. 75 .

As shown in Figure 2, let the midpoints of the six edges be $D$, $E$, $F$, $K$, $M$, $N$, and connect $B K$, $C K$.
Let $x=K D$, $y=M E$, $z=N F$.
By the median formula, we have
$$
\begin{aligned}
B K^{2} & =\frac{2 P B^{2}+2 A B^{2}-P A^{2}}{4}, \\
C K^{2} & =\frac{2 P C^{2}+2 C A^{2}-P A^{2}}{4}, \\
x^{2} & =\frac{2 B K^{2}+2 C K^{2}-B C^{2}}{4} \\
= & \frac{P B^{2}+P C^{2}+C A^{2}+A B^{2}-P A^{2}-B C^{2}}{4} .
\end{aligned}
$$

Similarly,
$$
\begin{array}{l}
y^{2}=\frac{P C^{2}+P A^{2}+A B^{2}+B C^{2}-P B^{2}-C A^{2}}{4}, \\
z^{2}=\frac{P A^{2}+P B^{2}+B C^{2}+C A^{2}-P C^{2}-A B^{2}}{4} .
\end{array}
$$

Therefore, $x^{2}+y^{2}+z^{2}$
$$
\begin{array}{l}
=\frac{P A^{2}+P B^{2}+P C^{2}+B C^{2}+C A^{2}+A B^{2}}{4} \\
=75 .
\end{array}
$$