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Let us be guided by the OP's comment: "I suggest to consider a majorizer "good" if it majorizes for $\sum x_i^2 = m$ but not for $\sum x_i^2 = m'$ for $m'>m$."

Let $\X_{n,m}$ be the set of all $X = (x_1, \ldots , x_n)$ such that $x_i$'s are nonnegative, $\sum_{i=1}^n x_i = n$, and $\sum_{i=1}^n x_i^2 = m$. Note that for $X\in\X_{n,m}$
\begin{equation*}
\sum x_i^2\le \Big(\sum x_i\Big)^2\le n\sum x_i^2,
\end{equation*}
so that
\begin{equation*}
n\le m\le m^2.
\end{equation*}
To avoid trivialities, suppose that $n\ge2$.
Let
\begin{equation*}
a_{n,m}:=\max\{x_1\colon X\in\X_{n,m}\}. \tag{1}
\end{equation*}
We shall see that
\begin{equation*}
a_{n,m}=1+\sqrt{(m-n)(1-1/n)}\in[1,n].
\end{equation*}
Let
\begin{equation*}
q:=\lfloor n/a_{n,m}\rfloor,\quad r:=n-qa_{n,m}\in[0,a_{n,m}),
\end{equation*}
\begin{equation*}
y_1=\cdots=y_q=a_{n,m},\quad y_{q+1}:=r,\quad y_i:=0\ \text{for }i=q+2,q+3,\dots.
\end{equation*}
Then clearly the vector $Y:=(y_1,\dots,y_n)$ majorizes all vectors in $\X_{n,m}$ -- but, for any $m'>m$, not all vectors in $\X_{n,m'}$ -- since $a_{n,m}$ is strictly increasing in $m$.

It remains to verify (1). The vector $X\in\X_{n,m}$ with the largest $x_1$ satisfies the Lagrange equations
\begin{equation}
1=\la+2\mu x_1,\quad 0=\la+2\mu x_i
\end{equation}
for some real $\la$ and $\mu$ and all $i\in J:=\{j\colon x_j>0\}$. If $\mu=0$, then $0=\la+2\mu x_i$ implies $\la=0$, which contradicts $1=\la+2\mu x_1$. So, $\mu\ne0$ and hence $x_1=a$ and $x_i=b$ for some positive real $a,b$ and all $i\in J$. So, letting $k$ stand for the cardinality of $J$, we have the system
of equations
\begin{equation}
a+kb=n,\quad a^2+kb^2=m.
\end{equation}
Solving this system, we get
\begin{equation}
a=\frac{n+\sqrt{k[(k+1) m-n^2]}}{k+1}
\end{equation}
if $(k+1)m\ge n^2$; otherwise, there is no solution. It is not hard to see that this expression for $a$ is increasing in $k\in[n^2/m-1,n-1]$, so that the maximum of this expression occurs at $k=n-1$. Thus, (1) is verified.

**Added in response to a comment by the OP:** Take any $X = (x_1, \ldots , x_n)\in\X_{n,m}$.

If the $x_i$'s are all the same, then they are all equal $1$, and hence $m=n$. So, in this case $\X_{n,m}$ is the singleton set $\{(1,\dots,1)\}$.

Consider now the nontrivial case when $n<m\le m^2$. Then the set $\X_{n,m}$ is a subset of the intersection of the $(n-1)$-dimensional sphere of radius $\sqrt m$ centered at the origin of $\R^n$ with the hyperplane in $\R^n$ given by equation $\sum_{i=1}^n x_i = n$, and this intersection is a $(n-2)$-dimensional sphere (say $S_{n,m}$) of radius $\sqrt{m-n}>0$. Suppose now a vector $Z=(z_1,\dots,z_n)\in\X_{n,m}$ majorizes a vector $X=(x_1,\dots,x_n)\in\X_{n,m}$. Then, by Rado's theorem, Theorem R, page 3266, $X$ is a convex combination of vectors obtained by permuting the coordinates $z_1,\dots,z_n$ of $Z$. If at least two coefficients of that convex combination are nonzero, then $X$ cannot belong to $S_{n,m}$, because all the points of a sphere are extreme points of the ball bounded by the sphere.

So, in any case, the only vectors in $\X_{n,m}$ that are majorized by a given vector $Z\in\X_{n,m}$ are the vectors obtained by permuting the coordinates $z_1,\dots,z_n$ of $Z$.

Thus, in the nontrivial case $n<m<m^2$, there is no vector in $\X_{n,m}$ that majorizes all vectors in $\X_{n,m}$ -- because in this case for any vector $Y\in\X_{n,m}$
there is another vector in $\X_{n,m}$ which cannot be obtained by permuting the coordinates of $Y$.