Orthogonal projections with $\sum P_i =I$, proving that $i\ne j \Rightarrow P_{j}P_{i}=0$

For each $j$, $$P_j=P_jIP_j=P_j\left(\sum_{k=1}^n P_k\right)P_j=\sum_{k=1}^nP_jP_kP_j=P_j+\sum_{k\neq j}P_jP_kP_j,$$ so $\sum\limits_{k\neq j}P_jP_kP_j=0$. For each $i\neq j$, $P_jP_iP_j=(P_iP_j)^*P_iP_j$ is a positive operator, and a sum of positive operators is positive, so $-P_jP_iP_j=\sum\limits_{k\neq i,j}P_jP_kP_j$ is also positive. This is only possible if $P_jP_iP_j=0$. Since $\|P_iP_j\|^2=\|(P_iP_j)^*P_iP_j\|=\|P_jP_iP_j\|$, it follows that $P_iP_j=0$.

Because of the properties you state, $\|P_{j}x\|^{2}=(x,P_{j}x)=(P_{j}x,x)$. Therefore, $$ \|x\|^{2} = (\sum_{j}P_{j}x,x)= \sum_{j}\|P_{j}x\|^{2}. $$ Apply this identity to $x=P_{k}y$, and use the fact that $P_{k}^{2}=P_{k}$: $$ \|P_{k}y\|^{2} = \sum_{j\ne k}\|P_{j}P_{k}y\|^{2}+\|P_{k}y\|^{2}. $$ The only way this can happen is $P_{j}P_{k}y=0$ for all $j \ne k$.

For all $i,j$ you have $P_i+P_j\le\sum_k P_k\le I$, hence $P_i\le I-P_j$.

Multiplying by $P_j$ on left and right on LHS and RHS you then get $$P_j P_i P_j\le P_j(I-P_j)P_j=0,$$ hence $P_j P_i P_j=0$, which implies $P_j P_i P_j=(P_j P_i)(P_j P_i)^\dagger=0$ and thus $P_i P_j=P_j P_i=0$.

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You can also prove the other direction: if $P_i P_j=0$ for all $i\neq j$ then $\sum_k P_k$ is a projector, as $$\left(\sum_k P_k\right)^2=\sum_k P_k + \sum_{i<j}(P_i P_j + P_j P_i)=\sum_k P_k.$$