Originally the problem had a typo, it missed the word "positive".

:Sigh: how many more typo's?

Anyone with a solution? Maybe Ankoganit with the official solution?

Take $m=1,n=31$ Then $b_1,b_2,\hdots,b_{31}=1,3,\hdots,61\implies b_1+b_2+\hdots+b_{31}=(1+61)\cdot\frac {31}{2}=961$ $a_1=2019-961=1058$ $5m+21n>581$

Indeed, the problem statement has a typo. The original quantity was $5m+12n$.

Redacted

As @above mentioned we have, $m(m+1)+n^2\le 2019$. or, $(m+\frac 12)^2+ n^2 \le 2019+ \frac 14$. All numbers $m,n$ that satisfy the above equation lie inside the circle $(x+\frac 12)^2 +y^2 \le 2019+\frac 14$ Just note that interesting points of the above circle and the straight line $5x+12y=581$ are $(13.617,42.74),(20.33,39.94)$(approx). Let's assume for the sake of contradiction that there exist integers $m,n$ which lie inside the circle but also satisfy $5m+12n >581$. Then this point must lie on the minor area bounded by the circle and the straight line. But observe that, for $n=40$,$m\le 19$,(since (m,n) must lie inside the circle) for $n=41$,$m\le 17$ for $n=42$,$m\le 15$ But each of these contradicts out assumption that $5m+12n >581$ Hence, $m(m+1)+n^2 \le 2019 \implies 5m+12n\le 581$ for integer $m,n$

Observe that $a_1 + a_2 + ... + a_m \ge 2 + 4 + .. + 2m = m(m+1) = m^2 + m$ Also, $b_1 + b_2 + ... + b_n \ge 1 + 3 + 5 + ... + 2n-1 = n^2$. So, we get that $n^2 + m^2 + m \le 2019$ Now, by Cauchy, we have that $(169)(2019) \ge 169(n^2 + m^2 + m) = 169(m^2 + n^2) + 169m = (5^2 + 12^2)(m^2 + n^2) + 169m \ge (5m+12n)^2 + 169m$ And so we have that $(5m+12n)^2 \le 169(2019-m) \implies 5m+12n \le 13 \sqrt{2019-m} \le 13 \sqrt{2019}$, which gives that $5m+12n \le 584$ Now, we just need to show that it isnt possible for $5m+12n$ to be any of $584,583,582$. For this, just find a general solution and use the fact that $m^2 + n^2 + m \le 2019$. Also, $n$ needs to be odd as otherwise the whole LHS would end up being even. So, since $5m+12n \le 584$ and we proved that it cant be any of $582,583,584$, we must have that $5m+12n \le 581$

clearly $a_{1}+a_{2}+\cdots+a_{m}+b_{1}+b_{2}+\cdots+b_{3}+\cdots+b_{n} \geqslant m(m+1)+n^2$ since all are distinct , now it implies $n^2+m^2+m \leqslant 2019 \implies n^2+m^2 \leqslant 2019-m$ , now $169(2019-m) \geqslant 169(m^2 + n^2)\stackrel{\text{C-S}}{\geqslant} (5m + 12n)^2 \implies 5m + 12n \leqslant 13\sqrt{2019-m} \leqslant 584$ We will show that $5m+12n=584,583,582$ is not possible. $5m+12n=584$ , clearly $n$ has to be odd and $n \equiv 2 \pmod 5$ which really doesn't give any such $m$ and $n$ to be working. $5m+12n=583$ $n \equiv 4 \pmod 5$ which also doesn't give any such $m$ and $n$ $5m+12n=582$ $n \equiv 1 \pmod 5$ which also doesn't works. Hence $5m+12n \leqslant 581$. $\blacksquare$

Observe that $m(m+1) + n^2 \le 2019 \implies (m+\frac 12) ^2 + n^2 \le \frac{8077}{4},$ so $$2025 >\frac{8077}{4} \ge (m+ \frac 12)^2 + n^2 \ge \frac{(5m + \frac 52 + 12n)^2}{13^2}$$$$\implies 5m + \frac 52 + 12n < 585 \implies 5m + 12n \le 582.$$Now show $5m + 12n \ne 582$ as above. $\square$