Let us prove by induction. For $n = 1$, we have to construct even numbers not exceeding $2 \cdot (4 + 1) = 10$.
$$\begin{aligned} +1 - 2 - 3 + 4 &= 0, \quad -1 + 2 - 3 + 4 &= 2, \quad +1 + 2 - 3 + 4 = 4, \\ +1 - 2 + 3 + 4 &= 6, \quad -1 + 2 + 3 + 4 = 8, \quad +1 + 2 + 3 + 4 = 10 \end{aligned}$$
By induction hypothesis assume that we've constructed all even numbers not exceeding $2n(4n + 1)$ by $\pm 1 \pm 2 \pm \cdots \pm 4n$.
Now let $m$ be even number, such that $2n(4n+1) < m \le (2n+2)(4n+5)$.
$$1 + 2 + \cdots + (4n + 4) = (2n + 2)(4n + 5)$$
$1+2+\cdots+(k-1)-k+(k+1)+\cdots+(4n+4) = (2n+2)(4n+5)-2k$, for $k = 1, 4n + 4$. The above expression gives us even numbers between $(2n + 2)(4n + 1)$ and $(2n + 2)(4n + 5)$.
Then $+1 + 2 + \cdots + (l - 1) - l + (l + 1) + \cdots + (4n + 3) - (4n + 4) = (2n + 2)(4n + 1) - 2l$ for $l = 1, 4n$. The above expression gives us even numbers between $2n(4n + 1)$ and $(2n + 2)(4n + 1)$.
For $m \le 2n(4n+1)$, we use $(4n+1)-(4n+2)-(4n+3)+(4n+4) = 0$ and induction hypothesis. By this the induction is complete.