Does $\sum_i A_i=I$ with $A_i$ positive imply $\{A_i\}_i$ are mutually diagonalisable? As discussed in this other question, if $A$ and $B$ are matrices such that $A+B=I$, then trivially they commute, and thus if they are both diagonalisable they are also mutually diagonalisable.
The same argument doesn't, however, apply when summing more than two such matrices. Suppose then that
$$\sum_{i=1}^n A_i = I.$$
The case of $A_i\ge0$ is the one I'm most interested about, but if positivity turns out to not be relevant for this, as it might very well be the case, feel free to weaken this constraint (to maybe consider Hermitian, normal, or just diagonalisable matrices).
If $\sum_i A_i=I$ then I can say that, for example, $[A_1,A_2+...+A_n]=0$, and thus $A_1$ and $\sum_{i>1} A_i$ are mutually diagonalisable. But then I cannot iterate the argument by splitting $A_2$ from $A_3+...+A_n$, as now they sum to a diagonal matrix (in their common eigenbasis), but not to the identity.
So does the result about mutual diagonalisability only work for $n=2$? A counterexample of three or more non-mutually-diagonalisable matrices summing to the identity would be a good answer.

It suffices to consider the case where $n=3$ if we put $0=A_4=A_5=\cdots$.
Let $A_1$ be any positive diagonal matrix whose diagonal entries are distinct and smaller than $1$. Let $B$ be any real symmetric matrix with a zero diagonal and nonzero off-diagonal entries. Then $A_2:=\frac12(I-A_1)+\epsilon B$ and $A_3:=\frac12(I-A_1)-\epsilon B$ are positive definite when $\epsilon$ is sufficiently small, but $A_1$ and $A_2$ are not simultaneously diagonalisable, because $A_1$ commutes only with diagonal matrices.
