Denote $s_n = a_n-a_{n-1}+a_{n-2}- \cdots$, thus $s_{n+1} = a_{n+1} - s_n$. Let $m\geq 1$ be the least number written on any card, hence $a_m > 0$, but $a_n = 0$ for all $0\leq n<m$, therefore $s_n = 0$ for all $0\leq n<m$ and $s_m = a_m>0$. Then $a_{m+1} - a_m = a_{m+1} - s_m = s_{m+1} \geq 0$, further on, $a_{m+2} - (a_{m+1} - a_m) = a_{m+2} - s_{m+1} = s_{m+2} \geq 0$, etc. Let $M\geq m$ be the largest number written on any card, hence $a_M > 0$, but $a_n = 0$ for all $n>M$. Then $0\leq s_{M+1} = a_{M+1} - s_M = -s_M$ forces $0\leq s_M \leq 0$, meaning $a_M - (a_{M-1} - a_{M-2} + \cdots + (-1)^{M-m+1} a_m) = a_M - s_{M-1} = s_M = 0$. It means we can pair $s_m$ cards bearing number $m$ with so many bearing number $m+1$, then $s_{m+1}$ cards bearing number $m+1$ with so many bearing number $m+2$, etc., until $s_{M-1}$ cards bearing number $M-1$ with so many bearing number $M$, and everything ends up well. Notice the irrelevancy of stating the number of cards is even; it follows to be even by the givens.