3. Let $a_{1}, a_{2}, \cdots, a_{100}$ be non-negative integers, satisfying the following conditions:
(1) There exists a positive integer $k \leqslant 100$, such that $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{k}$, and for $i>k$, $a_{i}=0$;
(2) $a_{1}+a_{2}+a_{3}+\cdots+a_{100}=100$;
(3) $a_{1}+2 a_{2}+3 a_{3}+\cdots+100 a_{100}=2022$.
Find the minimum possible value of $a_{1}+2^{2} a_{2}+3^{2} a_{3}+\cdots+100^{2} a_{100}$.
If $k \leqslant 20$, then $2022=a_{1}+2 a_{2}+\cdots+k a_{k} \leqslant k\left(a_{1}+a_{2}+\cdots+a_{k}\right)=100 k \leqslant 2000$, a contradiction, so $k \geqslant 21$.

If only one of $a_{1}, a_{2}, \cdots, a_{k}$ is greater than 0, then $a_{k}=100 \Rightarrow k a_{k} \neq 2022$, a contradiction;
If only two of $a_{1}, a_{2}, \cdots, a_{k}$ are greater than 0, then $a_{k-1}+a_{k}=100, (k-1) a_{k-1}+k a_{k}=(k-1)\left(a_{k-1}+a_{k}\right)+a_{k} \geqslant 2000+a_{k}$, and $a_{k} \geqslant \frac{1}{2}\left(a_{k-1}+a_{k}\right)=50$, a contradiction; If only three of $a_{1}, a_{2}, \cdots, a_{k}$ are greater than 0, then $a_{k-2}+a_{k-1}+a_{k}=100$,
$$
\begin{array}{l}
(k-2) a_{k-2}+(k-1) a_{k-1}+k a_{k}=(k-1)\left(a_{k-2}+a_{k-1}+a_{k}\right)+a_{k}-a_{k-2} \\
=100(k-1)+a_{k}-a_{k-2}=2022 \Rightarrow k=21, a_{k}-a_{k-2}=22 . \\
\text { Solving }\left\{\begin{array}{l}
a_{19}+a_{20}+a_{21}=100, \\
a_{21}-a_{19}=22
\end{array} \Rightarrow a_{19}=19+t, a_{20}=40-2 t, a_{21}=41+t(t \in \mathbf{Z}) .\right.
\end{array}
$$

Taking $a_{19}=19, a_{20}=40, a_{21}=41$, at this time $\sum_{i=1}^{k} i^{2} a_{i}=19^{2} \times 19+20^{2} \times 40+21^{2} \times 41$ $=20^{2}(19+40+41)+2 \times 20(41-19)+41+19=40000+880+60=40940$.

Below we prove that 40940 is the minimum value. If $a_{20} \leqslant 40$, then
$$
\begin{array}{l}
\sum_{i=1}^{k} i^{2} a_{i}=\sum_{i=1}^{k}[20-(20-i)]^{2} a_{i}=400 \sum_{i=1}^{k} a_{i}-40 \sum_{i=1}^{k}(20-i) a_{i}+\sum_{i=1}^{k}(20-i)^{2} a_{i} \\
=40000-80000+40 \times 2022+\sum_{i=1}^{k}(20-i)^{2} a_{i}=40880+\sum_{i=1}^{k}(20-i)^{2} a_{i} \\
\geqslant 40880+\sum_{\substack{1 \leqslant i \leqslant k \\
i \neq 20}} a_{i}=40880+100-a_{20} \geqslant 40880+60=40940 ;
\end{array}
$$

If $a_{20} \geqslant 41$, then $a_{21} \geqslant 42, \sum_{i=1}^{k} i^{2} a_{i}=\sum_{i=1}^{k}[19-(19-i)]^{2} a_{i}$
$$
\begin{array}{l}
=361 \sum_{i=1}^{k} a_{i}-38 \sum_{i=1}^{k}(19-i) a_{i}+\sum_{i=1}^{k}(19-i)^{2} a_{i} \\
=36100-2 \times 36100+38 \times 2022+\sum_{i=1}^{k}(19-i)^{2} a_{i}=40736+\sum_{i=1}^{k}(19-i)^{2} a_{i} \\
\geqslant 40736+a_{20}+4 a_{21} \geqslant 40736+41+4 \times 42=40945 .
\end{array}
$$

In summary, the minimum possible value of $a_{1}+2^{2} a_{2}+3^{2} a_{3}+\cdots+100^{2} a_{100}$ is 40940.