It gets a bit weird

$a_2=a_1+\color{#D61F06}{1}\\ a_3=a_1+\color{#D61F06}{2}\\$ and so on (as they are consecutive integers) $a_1^2+a_2^2+...+a_{19}^2=a_{20}^2+...+a_{37}^2$ $\Rightarrow a_1^2+(a_1+1)^2.....+(a_{18}+1)^2=(a_{19}+1)^2+(a_{20}+1)^2....(a_{36}+1)^2$ $\Rightarrow a_1^2+2a_1(1+2..+18-19-20....-36)+(1^2+2^2+....18^2-19^2-20^2-....-36^2)=0$ $\Rightarrow a_1^2+2a_1(\color{#20A900}{2(\displaystyle \sum_{i=1}^{18} i)-(\displaystyle \sum_{i=1}^{36}i)})+(\color{#3D99F6}{2(\displaystyle \sum_{i=1}^{18}i^2)-(\displaystyle \sum_{i=1}^{36}i^2)})\\$ $=a_1^2-648a-11988=0$ $a_1=-18,666$

Largest possible value of $a_1$ = $\color{magenta}{666}\\$

I took advantage of the near-symmetry of the equation and set $a_{19} =: n$ , so that $n^2 + \sum_{k = 1}^{18} (n-k)^2 = \sum_{k=1}^{18} (n+k)^2.$ In $(n\pm k)^2 = n^2 + k^2 \pm 2 nk$ , the squares cancel each other on both sides so that we simply get $n^2 = \sum_{k=1}^{18} 4nk.$ One obvious solution is $n = 0$ , which would give $a_1 = -18$ . The other solution is positive: $n = 4\sum_{k=1}^{18} k = 4\cdot \frac{19\cdot 18}2 = 684.$ Finally, $a_1 = a_{19} - 18 = n - 18 = \boxed{666}$ .