$\newcommand\la\lambda\newcommand\ep\varepsilon\newcommand\ip[2]{\langle #1,#2\rangle}\newcommand\Span{\operatorname{span}}$As noted in a comment by Yemon Choi, the question is not well posed.

Apparently, the OP wanted to have a *good*, nontrivial sufficient condition, preferably close to necessity. Such a sufficient condition will be provided here, and the sufficient condition will also be necessary when $E$ is finite dimensional.

Let
\begin{equation}
U:=A^*A=\int_{[0,\|A\|^2]}\la\,dP^A_\la\quad\text{and}\quad
V:=B^*B=\int_{[0,\|B\|^2]}\la\,dP^B_\la \tag{0}\label{0}
\end{equation}
be the spectral decompositions of the self-adjoint bounded linear operators $A^*A$ and $B^*B$. (Note that the families $(P^A_\la)$ and $(P^B_\la)$ of orthoprojectors are uniquely determined by condition \eqref{0}.)

Let
$$Q^A_\ep:=\int_{[\|A\|^2-\ep,\|A\|^2]}dP^A_\la, \quad Q^B_\ep:=\int_{[\|B\|^2-\ep,\|B\|^2]}dP^B_\la,$$
$$E^A_\ep:=Q^A_\ep E,\quad
E^B_\ep:=Q^B_\ep E. $$

Then the condition
\begin{equation}
E^A_\ep\cap E^B_\ep\ne\{0\}\quad\forall\ep>0 \tag{1}\label{1}
\end{equation}
is sufficient for $N_1(A,B)=N_2(A,B)$.

Indeed, if \eqref{1} holds, then for each real $\ep>0$ there is a unit vector
$$u_\ep\in E^A_\ep\cap E^B_\ep,$$
so that $Q^A_\ep u_\ep=u_\ep=Q^B_\ep u_\ep$.
So,
$$\begin{aligned}
\|Au_\ep\|^2&=\ip{A^*Au_\ep}{u_\ep} \\
&=\int_{[0,\|A\|^2]}\la\,d\ip{P^A_\la u_\ep}{u_\ep} \\
&=\int_{[0,\|A\|^2]}\la\,d\ip{P^A_\la Q^A_\ep u_\ep}{Q^A_\ep u_\ep} \\
&=\int_{[\|A\|^2-\ep,\|A\|^2]}\la\,d\ip{P^A_\la Q^A_\ep u_\ep}{Q^A_\ep u_\ep} \\
&\ge(\|A\|^2-\ep)\int_{[\|A\|^2-\ep,\|A\|^2]}d\ip{P^A_\la Q^A_\ep u_\ep}{Q^A_\ep u_\ep} \\
&=(\|A\|^2-\ep)\ip{Q^A_\ep u_\ep}{Q^A_\ep u_\ep} \\
&=(\|A\|^2-\ep)\ip{u_\ep}{u_\ep}
=\|A\|^2-\ep.
\end{aligned}$$
Similarly, $\|Bu_\ep\|^2\ge\|B\|^2-\ep$. So,
$$N_1(A,B)^2\ge\|Au_\ep\|^2+\|Bu_\ep\|^2
\ge\|A\|^2+\|B\|^2-2\ep=N_2(A,B)^2-2\ep$$
for any real $\ep>0$.
Thus, $N_1(A,B)\ge N_2(A,B)\ge N_1(A,B)$. $\quad\Box$

*Remark 1:* Condition \eqref{1} is necessary for $N_1(A,B)=N_2(A,B)$ when $E$ is finite dimensional. Indeed, suppose that $N_1(A,B)=N_2(A,B)$. Then for some sequence $(x_n)$ on the unit sphere $S$ in $E$ we have
\begin{equation}
\|A\|^2+\|B\|^2\ge\|Ax_n\|^2+\|Bx_n\|^2\to\|A\|^2+\|B\|^2.
\end{equation}
It follows that $\|Ax_n\|^2\to\|A\|^2$ and $\|Bx_n\|^2\to\|B\|^2$. The unit sphere $S$ in the finite-dimensional Hilbert space $E$ is compact. So, passing to a subsequence, without loss of generality we have $x_n\to x$ for some $x\in S$. So, $\|Ax\|^2=\|A\|^2$ and $\|Bx\|^2=\|B\|^2$, which implies \eqref{1}. $\quad\Box$

*Remark 2:* Condition \eqref{1} is not in general necessary for $N_1(A,B)=N_2(A,B)$ when $E$ is infinite dimensional. E.g., let $E:=\ell^2$, with the standard basis $(e_1,e_2,\dots)$.

Let $U$ be the positive-semidefinite self-adjoint linear operator on $E$ whose matrix (in the standard basis) is the block-diagonal matrix with the $2\times2$ diagonal blocks
\begin{equation}
D_j^U:=\frac j{j+1}\,P_j^U,\quad\text{where}\quad
P_j^U:=\begin{bmatrix}1&0\\0&0 \end{bmatrix}
\end{equation}
for integers $j\ge1$.
The factor $\frac j{j+1}$, strictly increasing to $1$ in $j$, was introduced to force gliding to $\infty$ in search of a (nonexistent) maximizer $x\in S$ of $\ip{Ux}x$.

Similarly, let $V$ be the positive-semidefinite self-adjoint linear operator on $E$ whose matrix (in the standard basis) is the block-diagonal matrix with the $2\times2$ diagonal blocks
\begin{equation}
D_j^V:=\frac j{j+1}\,P_j^V,\quad\text{where}\quad
P_j^V:=\frac1{j^2+1}\,\begin{bmatrix}j^2&j\\j&1 \end{bmatrix}
\end{equation}
for integers $j\ge1$.
Note that $P_j^U$ and $P_j^V$ are orthoprojector matrices of rank $1$ each, with the respective column spaces $\Span\Big\{\begin{bmatrix}1\\0 \end{bmatrix}\Big\}$ and $\Span\Big\{\begin{bmatrix}j\\1 \end{bmatrix}\Big\}$.

In accordance with \eqref{0}, let $A:=\sqrt{U}$ and $B:=\sqrt{V}$. Then $\|A\|^2=\|U\|=1$ and $\|B\|^2=\|V\|=1$. Moreover, letting an integer $i$ go to $\infty$, we get $$N_1(A,B)\ge\|Ae_{2i}\|^2+\|Be_{2i}\|^2=\ip{Ue_{2i}}{e_{2i}}+\ip{Ve_{2i}}{e_{2i}} \\ \to1+1=\|A\|^2+\|B\|^2=N_2(A,B),$$
which implies that $N_1(A,B)=N_2(A,B)$.

However, for any integer $j\ge1$ and any $\ep\in(\frac1{j+2},\frac1{j+1}]$,
\begin{equation}
E_\ep^A=\Span\{e_{2i}\colon i\ge j\}\quad\text{and}\quad
E_\ep^B=\Span\{e_{2i}+\tfrac1{2i+1}\,e_{2i+1}\colon i\ge j\},
\end{equation}
so that \eqref{1} fails to hold. $\quad\Box$

It appears that the following condition is necessary and sufficient for $N_1(A,B)=N_2(A,B)$:

There exist sequences $(a_n)$ and $(b_n)$ such that $a_n-b_n\to0$,
$a_n\in S\cap E_{1/n}^A$ for all $n$, and $b_n\in S\cap E_{1/n}^B$ for all $n$.

This conjecture is probably easy to prove, but ....

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