Does no one have solution to this problem ?

How could one have a solution, when one of the condition is this, $\sum_{i=1}^{n}=12$, missing the summand?

May I supose that the sum is: $a_1+a_2+ \dots +a_n=12$

Can this be trusted? See Problem No.3.

(I assume that 0 is a natural number.) The following settles all cases except $n=4$: Then $n=1$ does not work, since $a_1=12$ is not square-free. $n=2$ works with $m=98$, and $m+1=99$ and $m+2=100$; (and $a_1+a_2=11+1=12$). $n=3$ works with $m=3$ (and $a_1+a_2+a_3=1+5+6$). $n=5$ works with $m=0$ (and $a_1+a_2+a_3+a_4+a_5=1+2+3+1+5$). Finally $n\ge6$ is impossible: the cases with $m\le3$ do not work. For $m\ge4$ at most one of the numbers $m+1,m+2,\ldots,m+6$ is a perfect square with $a_i=1$. Every odd non-square has $a_i\ge3$, and every even non-square has $a_i\ge2$; this brings the sum above $12$.

(In Bulgaria $0$ is not a natural number.) We cannot have $n=1$ since $12$ is not squarefree, we cannot have $n\geq 13$ and we cannot have $8 \leq n \leq 12$ since among $12$ consecutive integers at most $3$ are perfect squares (the sequence of squares is $1$, $4$, $9$, $16$, $\ldots$ and have in mind the increasing gaps) and hence at least $n-3$ of the $a_i$-s are at least $2$, giving the bound $\sum a_i \geq 2(n-3) + 3 = 2n - 3 \geq 13 > 12$. Hence $n\leq 7$. The very helpful observation which reduces the case bash significantly is that all $a_i$-s are distinct! Indeed, if not, then $6 \geq n - 1 \geq (m + j) - (m + i) = a(b_j^2 - b_i^2) \geq 3a$, forcing $a=2$, $b_j = 2$, $b_i = 1$ (and hence a sequence containing $2$ and $8$, so $2$, $3$, $4$, $5$, $6$, $7$, $8$, - but the sum of the squarefree parts exceeds $12$) or $a=1$ (so $b_i \neq 1$ since $m+i > 1$), $b_j = 3$, $b_i = 2$ (and so a sequence with $6$ or $7$ numbers containing $4$, $5$, $6$, $7$, $8$, $9$, again with sum of squarefree parts more than $12$). In particular, since each $a_i$ is $1$, $2$, $3$, $5$, $6$ or more and $1+2+3+5+6 > 12$, we must have $n\leq 4$. If $n=2$ then a possibility is $a_1 = 1$, $a_2 = 11$ and we get an example with $m=98$ (the numbers $99 = 11 \cdot 3^2$ and $100 = 1 \cdot 10^2$). If $n=3$ then a possibility is $a_1 = 1$, $a_2 = 5$, $a_3 = 6$ and we get an example with $m=3$ (the numbers $4 = 1 \cdot 2^2$, $5 = 5 \cdot 1^2$, $6 = 6 \cdot 1^2$). Finally, if $n=4$, the only possibillity is $a_1, a_2, a_3, a_4$ to be $1,2,3,6$ in some order and so $(m+1)(m+2)(m+3)(m+4) = (6K)^2$ for some integer $K$, but the left hand side is $(m^2+5m+4)(m^2+5m+6) = (m^2 + 5m + 5)^2 - 1$ and is hence not a square. Therefore the only possible $n$ are $2$ and $3$.