Call an $n$-tuple $(a_1, \dots, a_n)$ nice if it satisfies all the conditions. Note that scaling by a positive constant does not affect the niceness of an $n$-tuple. WLOG assume $a_1+\dots+a_n=\tfrac{n}{2}$. Note that $a_1<0$ and $a_n>0$ so we can find an $1<\ell<n$ with $a_1 \le \dots \le a_{\ell} \le 0<a_{\ell+1}\le \dots \le a_n$. Suppose $2\ell>n$. Now define $X \overset{\text{def}}{:=} a_{\ell+1}+\dots+a_n$ and $Y \overset{\text{def}}{:=} -(a_1+\dots+a_{\ell})$. Note $X,Y$ are positive reals and $2(X-Y)=n>0$. By Cauchy-Schwartz \begin{align*} \frac{n}{2}=\frac{2}{n} \cdot \left(\frac{n}{2}\right)^2&=\frac{2}{n}(a_1+\dots+a_n)^2 \\ &= \left((-a_1)^2+\dots+(-a_{\ell})^2\right)+\left(a_{\ell+1}^2+\dots+a_n^2\right) \\ & \ge \frac{Y^2}{\ell}+\frac{X^2}{n-\ell} \end{align*}Now we show $$\frac{X^2}{n-\ell}+\frac{Y^2}{{\ell}} \ge \frac{2(X^2+Y^2)}{n}.$$Suppose this is not the case; then $$X^2\left(\frac{1}{n-\ell}-\frac{2}{n}\right) \le Y^2\left(\frac{2}{n}-\frac{1}{\ell}\right) \implies 1 \le \frac{X^2}{Y^2} \le \frac{n-\ell}{\ell}$$(since $2\ell-n>0$ is cancelled) hence $2\ell \le n$; a contradiction! Then $$(X-Y)^2=\left(\frac{n}{2}\right) \ge X^2+Y^2$$by the previous observation. Thus, $XY \le 0$; which is false since $X,Y>0$. We conclude that $m>n/2 \implies a_m>0$ as desired.

Assume for the sake of contradiction that $a_m\le 0$. Then we must have $a_1\le a_2\le \cdots\le a_m\le 0$. Observe that \begin{align*} 0 &\ge a_1+a_2+\cdots+a_m \\ a_n+a_{n-1}+\cdots+a_{m+1} &\ge a_1+a_2+\cdots+a_{n-1}+a_n. \\ \end{align*}Since $a_1+a_2+\cdots+a_n>0$ we get $$(a_n+a_{n-1}+\cdots+a_{m+1})^2\ge (a_1^2+a_2^2+\cdots+a_n^2)=\dfrac{n}{2}(a_1^2+a_2^2+\cdots+a_n^2).$$By Titu's Lemma we have $(n-m)(a_n^2+a_{n-1}^2\cdots+a_{m+1}^2)\ge (a_n+a_{n-1}+\cdots+a_{m+1})^2$. Hence we obtain \begin{align*} (n-m)(a_n^2+a_{n-1}^2+\cdots+a_{m+1}^2) &\ge \dfrac{n}{2}(a_n^2+a_{n-1}^2+\cdots +a_1^2) \\ \left(\dfrac{n}{2}-m\right)(a_n^2+a_{n-1}^2+\cdots+a_{m+1}^2) &\ge \dfrac{n}{2}(a_m^2+a_{m-1}^2+\cdots+a_1^2)\ge0. \end{align*} But we have reached a contradiction since $\left(\dfrac{n}{2}-m\right)(a_n^2+a_{n-1}^2+\cdots+a_{m+1}^2)<0$ as $m=\lfloor n/2\rfloor+1$. Therefore we must have $a_m>0$ as desired.

Lemma:$n(a_1^2+a_2^2+...+a_n^2) \ge (a_1+a_2+...+a_n)^2$ for all real numbers $a_1,a_2,...,a_n$. Proof:With Titu Lemma. NOTE:In this question equality is not hold.Because,otherwise all numbers are equal and positive. First case: $n=2k$. $$2(a_1+a_2+...+a_{2k})^2=2k(a_1^2+a_2^2+...+a_{2k}^2)$$$$(a_1+a_2+...a_k)^2+(a_{k+1}+a_{k+2}+...+a_{2k})^2+2(a_1+a_2+...+a_{k})(a_{k+1}+a_{k+2}+...+a_{2k})=k(a_1^2+a_2^2+...+a_{k}^2)+k(a_1^2+a_2^2+...+a_{2k}^2)$$$$(a_1+a_2+...a_k)^2+(a_{k+1}+a_{k+2}+...+a_{2k})^2+2(a_1+a_2+...+a_{k})(a_{k+1}+a_{k+2}+...+a_{2k}) > (a_1+a_2+...a_k)^2+(a_{k+1}+a_{k+2}+...+a_{2k})^2$$$$2(a_1+a_2+...+a_{k})(a_{k+1}+a_{k+2}+...+a_{2k}) > 0$$Subcase 1:$a_1+a_2+...+a_{k}$ and $a_{k+1}+a_{k+2}+...+a_{2k}$ are  negative.Then their sum is also negative. But their sum must be positive. Subcase 2:$a_1+a_2+...+a_{k}$ and $a_{k+1}+a_{k+2}+...+a_{2k}$ are positive.We have exist $i$ such that $a_i$ is positive,where $i=1,2,...,k$.So,$a_j$ is positive for all $2k \ge j>i$.So,$a_{k+1}>0$. Second case : $n=2k+1$. Assume $a_{k+1}<0$ $$2(a_1+a_2+...+a_{2k+1})^2=(2k+1)(a_1^2+a_2^2+...+a_{2k+1}^2)$$$$2(a_1+a_2+...a_k)^2+2(a_{k+2}+a_{k+3}+...+a_{2k+1})^2+4(a_1+a_2+...+a_{k})(a_{k+2}+a_{k+3}+...+a_{2k+1})+2a_{k+1}(a_1+a_2+...+a_{2k+1})=2k(a_1^2+a_2^2+...+a_{k}^2)+2k(a_{k+2}^2+a_{k+3}^2+...+a_{2k+1}^2)+a_1^2+a_2^2+...+a_{k}^2+a_{k+2}^2+a_{2k+1}^2+(2k+1)a_{k+1}^2$$ We use lemma here. $$ 4(a_1+a_2+...+a_{k})(a_{k+2}+a_{k+3}+...+a_{2k})=a_1^2+a_2^2+...+a_{k}^2+a_{k+2}^2+a_{2k+1}^2+(2k+1)a_{k+1}^2-2a_{k+1}(a_1+a_2+...+a_{2k+1})>0$$And we know $0>a_{k+1} \ge a_k \ge ... \ge a_1$. So,$a_1+a_2+...+a_k$ is negative. So,$a_{k+2}+a_{k+3}+...+a_{2k+1}$ is negative.But this is not true.Because, their sum is must be positive.So,our assume is not true.So,$a_{k+1}>0$.

Here's a totally different solution which doesn't use any inequality: Note that the given condition is conserved under the transformation $a_i \rightarrow \lambda a_i$. So WLOG we can assume $$a_1+a_2+ \dots +a_n=\frac{n}{2} \Longrightarrow a_1^2+a_2^2+ \dots +a_n^2=\frac{n}{2}$$Define $b_i=1-a_i$ for all $i \in \{1,2, \dots ,n\}$. Then we get $$b_1^2+b_2^2+ \dots +b_n^2=\sum_{i=1}^n (1-a_i)^2=n+\sum_{i=1}^n (a_i^2-2a_i)=\sum_{i=1}^n a_i^2=\frac{n}{2}$$Now, As $<a_i>$ is a non-decreasing sequence, so $<b_i>$ must be a non-increasing sequence. We claim that $b_m<1$. FTSOC assume that $b_m \geq 1$. Then $b_1 \geq b_2 \geq \dots \geq b_m \geq 1$. Thus, we have $$\sum_{i=1}^n b_i^2 \geq m+\sum_{i=m+1}^n b_i^2 \geq m \Rightarrow \frac{n}{2} \geq \lfloor n/2\rfloor+1 \rightarrow \text{CONTRADICTION}$$This means that $1-a_m=b_m<1$, i.e. $a_m>0$. Hence, done. $\blacksquare$