Note that $(3n)^2 + (4n)^2 = (5n^2)$ from where we derive $(3n)^2 + (4n - 1)^2 - (5n - 1)^2 = 2n$ and $(3n + 2)^2 + (4n)^2 - (5n - 1)^2 = 2n + 3$.

a. Let's consider the case of quadratical number is odd. Set $a = 3n + 2$, $b = 4n$, $c = 5n + 1$ in the equality $(3n + 2)^2 + (4n)^2 - (5n - 1)^2 = 2n + 3$. If $n \ge 3$ then the condition $a < b < c$ holds. Therefore if $m \ge 9$ then any odd number $m = 2n + 3$ can be written in the form $m = 2n + 3 = a^2 + b^2 + c^2$ as noted above. Now consider the cases $m = 1, 3, 5, 7$.

$1 = 4^2 + 7^2 - 8^2$, $3 = 4^2 + 6^2 - 7^2$, $5 = 4^2 + 5^2 - 6^2$, $7 = 10^2 + 14^2 - 17^2$. Thus every odd number is quadratical.

b. Let's consider the case of quadratical number is even. Set $a = 3n$, $b = 4n - 1$, $c = 5n - 1$ in the equality $(3n)^2 + (4n - 1)^2 - (5n - 1)^2 = 2n$. If $n \ge 2$ then the condition $a < b < c$ holds. Note that $2 = 5^2 + 11^2 - 12^2$ and from where we conclude that every even number is quadratical.

Hence all numbers 1 - 2014 are quadratical.