Let $\{Y_i\}_{i=1}^N\in\mathbb{R}^{n\times m}$ be a set of full column rank matrices (i.e., $\mathrm{rank}(Y_i)=m$ for all $i$) and $\{P_i\}_{i=1}^N\in\mathbb{R}^{m\times m}$ be a set of positive definite matrices. Let $A\in\mathbb{R}^{m\times n}$, $B\in\mathbb{R}^{m\times n}$ and consider the following equation $$\tag{1}\label{eq:1} \sum_{i=1}^N Y_i Y_i^\top \Delta_i Y_i Y_i^\top=0_{n\times n}, $$ where $$ \Delta_i := A^\top P_i B+B^\top P_i A, $$ $0_{n\times n}$ denotes the $n\times n$ zero matrix and $(\cdot)^\top$ denotes transposition.
My question. Does \eqref{eq:1} imply $Y_i^\top \Delta_i Y_i=0_{m\times m}$ for all $i=1,2,\dots,N$?
A simple observation. The answer is in the affirmative if $P_i=p_i M$, for all $i$, with $M\in\mathbb{R}^{m\times m}$ being a positive definite matrix and $\{p_i\}_{i=1}^N$ being a set of positive real numbers. Indeed, this follows, for instance, from the fact that \eqref{eq:1} is equivalent to $$\tag{2}\label{eq:2} \sum_{i=1}^N \mathrm{tr}(Y_i^\top XY_i Y_i^\top \Delta_i Y_i) = 0, \ \ \text{for all symmetric } X\in\mathbb{R}^{n\times n}, $$ where $\mathrm{tr}(\cdot)$ denotes the trace operator. More precisely, if we pick $X=A^\top MB+B^\top M A$ in the previous expression, the LHS of \eqref{eq:2} turns into a sum positive numbers which implies that $Y_i^\top \Delta_i Y_i=0_{m\times m}$ for all $i=1,2,\dots,N$, as desired.
Concerning the case of arbitrary positive definite $\{P_i\}_{i=1}^N$, so far I didn't manage to find neither a proof or counterexample to my question. So any help would be really appreciated! Thanks in advance!
For $N=1$, the result is true: if $YY^\top\Delta YY^\top=0$, then multiplying by $Y^\top$ on the left and by $Y$ on the right we get $$ (Y^\top Y)Y^\top \Delta Y(Y^\top Y)=0. $$ As $Y^\top Y$ is invertible, it follows that $Y^\top \Delta Y=0$.
For $N\geq2$, the result is false. Let $n=m=2$, and $$ Y_1=Y_2=I_2,\ P_1=\begin{bmatrix} 2&1\\1&2\end{bmatrix},\ P_2=\begin{bmatrix} 2&-1\\-1&2\end{bmatrix},\ A=\begin{bmatrix} 0&0\\1&0 \end{bmatrix},\ B=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$ Then $$ \Delta_1=A^\top P_1B+B^\top P_1 A=\begin{bmatrix}2&0\\0&0\end{bmatrix},\ \ \Delta_2=A^\top P_2B+B^\top P_2 A=\begin{bmatrix}-2&0\\0&0\end{bmatrix}. $$ So $$ Y_1Y_1^\top\Delta_1Y_1Y_1^\top+Y_1Y_2^\top\Delta_2Y_2Y_2^\top=\Delta_1+\Delta_2=0, $$ while $$ Y_1^\top\Delta_1 Y_1=\Delta_1\ne0,\ \ Y_2^\top\Delta_2 Y_2=\Delta_2\ne0. $$
The result, on the other hand, is true when $\Delta_j$ is positive for all $j$.