Suppose $A$ and $B$ are positive operators on a finite-dimensional inner product space $V$ and $\|A + B\| = \|A\| + \|B\|$.
Because $A$ and $B$ are positive, $A+B$ is also a positive operator. Thus there exists $x \in V$ such that$$\|x\| = 1 \quad \text{and} \quad\|A+B\| = \langle (A+B)x, x \rangle.
$$
Now
\begin{align*}
\|A+B\| &= \langle (A+B)x, x \rangle\\[3pt]
&= \langle Ax, x \rangle + \langle Bx, x \rangle\\[3pt]
&\le \|A\| + \|B\|\\[3pt]
& = \|A+B\|.
\end{align*}
Because the first term above equals the last term above, the inequality must be an equality. In other words,
$$
\langle Ax, x \rangle = \|A\| \quad \text{and} \quad \langle Bx, x \rangle = \|B\|.
$$
The equations above imply that $x$ in an eigenvector of $A$ and $x$ is an eigenvector of $B$. Hence we have shown that $A$ and $B$ have a common eigenvector, as desired.