Here, page 47, the solution was published. I haven't found the Statement yet.
We select the property in the photo.
ANSWER CiP
We will show that one of the numbers $a_k$ is equal to 1,
and the others are zero. Then $b=k$.
Solution CiP
We will treat the case $n=4$ in detail, so for
$a_1,\;\;a_2,\;\;a_3,\;\;a_4\geqslant 0 \tag{1}$
we have equations
\begin{cases}a_1+a_2+a_3+a_4=1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)\\1\cdot a_1+2\cdot a_2+3\cdot a_3+4\cdot a_4=b\;\;\;\;\;\;\;\;\;\;\;\;(3)\\1^2\cdot a_1+2^2\cdot a_2+3^2\cdot a_3+4^2\cdot a_4=b^2\;\;\;\;\;(4)\end{cases}
Subtracting (2) from (3) we have $b-1=a_2+2\cdot a_3+3\cdot a_4\geqslant 0$, so $b\geqslant 1.$
$\textbf{If}\;\;\bf{b\geqslant 4}$ then
$b^2-4\cdot b\;\underset{(4)-4\times (3)}{=}(a_1+4a_2+9a_3+16a_4)-4\cdot (a_1+2a_2+3a_3+4a_4)$, so
$0\leqslant b^2-4b=-3a_1-4a_2-3a_3\overset{(1)}{\leqslant} 0 \tag{5}$
Therefore, in (5) we have everywhere the sign $"="$. In particular
$b=4,\;\;a_1=a_2=a_3=0,\;\;and\;\;a_4\overset{(2)}{=}1 \tag{6}$
$\textbf{Let}\;\;\bf{b\in[i,i+1)}$ for some $i\in\{1,\;2,\;3\}$. We calculate first
$b^2-b\cdot i\;\overset{(4)}{\underset{i\times (3)}{=}}(1a_1+4a_2+9a_3+16a_4)-(i\cdot a_1+2i\cdot a_2+3i\cdot a_3+4i\cdot a_4)$ so
$b^2-bi=(1-i)a_1+2(2-i)a_2+3(3-i)a_3+4(4-i)a_4 \tag{7}$
On the other hand
$b-i\cdot 1\overset{(3)}{\underset{i\times (2)}{=}}(a_1+2a_2+3a_3+4a_4)-(i\cdot a_1+i\cdot a_2+i\cdot a_3+i\cdot a_4)=$
$=(1-i)a_1+(2-i)a_2+(3-i)a_3+(4-i)a_4$
which, multiplied by $b$ gives us
$b^2-bi=b(1-i)a_1+b(2-i)a_2+b(3-i)a_3+b(4-i)a_4 \tag{8}$
Equating (7) and (8)
$(1-i)a_1+2(2-i)a_2+3(3-i)a_3+4(4-i)a_4=b(1-i)a_1+b(2-i)a_2+b(3-i)a_3+b(4-i)a_4$
we obtain
$(b-1)(i-1)a_1+(b-2)(i-2)a_2+(b-3)(i-3)a_3+(b-4)(i-4)a_4=0 \tag{9}$
For $i=1$, (so $1\leqslant b<2$), the first term cancels out, and all coefficients of $a_2,\;a_3,\;a_4$ are $>0$.
Therefore $a_2=a_3=a_4=0$, so (cf. (2)) $a_1=1$ and then $b\overset{(3)}{=}1$.
For $i=2$, (so $2\leqslant b<3$), then the second term cancels out, and all coefficients of $a_1,\;a_3,\;a_4$ are $>0$.
Therefore $a_1=a_3=a_4=0$, so (cf.(2)) $a_2=1$ and then $b\overset{(3)}{=}2$.
For $i=3$, (so $3\leqslant b <4$), then the third term cancels out, and all coefficients of $a_1,\;a_2,\;a_4$ are $>0$.
Therefore $a_1=a_2=a_4=0$, so (cf.(2)) $a_3=1$ and then $b\overset{(3)}{=}3$.
The last three conclusions, together with (6), validate the answer in this particular case.
The general case extends what is shown in case $n=4$. So let be the numbers
$a_1,\;a_2,\;\dots,a_n\;\geqslant 0 \tag{11}$
that verify the relations
$$a_1+a_2+\dots+a_n=\sum_{k=1}^na_k=1 \tag{12}$$
$$a_1+2\cdot a_2+\dots+n\cdot a_n=\sum_{k=1}^nk\cdot a_k=b \tag{13}$$
$$a_1+4\cdot a_2+\dots +n^2\cdot a_n=\sum_{k=1}^nk^2\cdot a_k=b^2 \tag{14}$$
We have
$$b-1\overset{(13),(12)}{=}\sum_{k=1}^nk\cdot a_k-\sum_{k=1}^na_k=\sum_{k=1}^n(k-1)\cdot a_k=\sum_{k=2}^n(k-1)a_k\geqslant 0$$
so $b\geqslant 1$.
If $b\geqslant n$ then
$$0\leqslant b^2-n\cdot b\overset{(14),(13)}{=}\sum_{k=1}^nk^2\cdot a_k-n \sum_{k=1}^nk\cdot a_k=\sum_{k=1}^n(k^2-nk)a_k=\sum_{k=1}^{n-1}k(k-n)a_k\leqslant 0$$
and according to (11) it result $a_1=\dots=a_{n-1}=0,\;b=n,\;a_n\overset{(12)}{=}1$.
If $b\in[i,i+1)$ for some $i\in\{1,\dots,n-1\}$, then, on the one hand
$$b^2-i\cdot b\;\overset{(14),(13)}{=}\sum_{k=1}^nk^2a_k-i\cdot \sum_{k=1}^nka_k=\sum_{k=1}^nk(k-i)a_k \tag{15}$$
On the other hand
$$b-i\;\overset{(13),(12)}{=}\sum_{k=1}^nka_k-i\cdot \sum_{k=1}^na_k=\sum_{k=1}^n(k-i)a_k$$
which, multiplied by $b$ gives us
$$b^2-ib=\sum_{k=1}^nb(k-i)a_k \tag{16}$$
Equating (15) and (16) we obtain
$$\sum_{k=1}^n(b-k)(i-k)a_k=0$$
or, excluding the term $a_i$ wich has coefficient zero
$(b-1)(i-1)a_1+\dots+(b-i+1)a_{i-1}+(b-i-1)(-1)a_{i+1}+\dots (b-n)(i-n)a_n=0$.
Let's analyze the coefficients of the numbers $a_k$ in the equation above :
for $k<i\;\Rightarrow\;(b-k)(i-k)\;\overset{k<i\leqslant b}{>}0;$
for $k>i\;\Rightarrow\;(b-k)(i-k)=(k-b)(k-i)\overset{k\geqslant i+1>b}{>}0$
Then it turns out that $a_1=\dots=a_{i-1}=0=a_{i+1}=\dots =a_n$, so $a_i\overset{(12)}{=}1$ and $b\underset{(13)}{=}i.$
$\blacksquare$
