This might be anything from trivial to not trivial, but my probability theory is too rusty to get even started with this.
Supppose I have $N$ real numbers $(x_1, x_2, \ldots, x_N)$, each sampled independently from $\mathcal{N}(0,1)$. Conditional on their sum being $0$, how could I calculate the probability that for a given $k < N$ and $\varepsilon > 0$ the sum $\sum_{j=1}^k x_j$ is in $[-\varepsilon,\varepsilon]$?
Surely the probability will be much better than without the conditional part? Without it we'd just look at the $k$-fold sum of independent normal distributions, i.e. $\mathcal{N}(0,k)$, but here the condition that we get to $0$ at $N$ should surely limit our variations? Some simple simulations seem to support this as well.
Bonus points if your solution can help me to also work with other base distributions besides $\mathcal{N}(0,1)$.
Recall that if $(X,Y)\sim N(m_X,m_Y, \Sigma)$ with $$\Sigma=\left[\begin{array}{cc}A&B^T\\B&C\end{array}\right]$$ (with obvious notations) then the conditional law of $X|Y$ has mean $m_X+B^TC^{-1}(Y-m_Y)$ and a covariance independent of $Y$ equal to $$A-B^TC^{-1}B$$ Let us apply this to $$X=(X_1,\ldots,X_n), \ \ Y=X_1+\cdots+X_n$$ where the iid $X_i$ are $N(0,1).$ Then $A_n=I_n$, $C_n=n$ and $B_n=(1,\ldots,1).$ We get $$X|Y=0\sim N(0,I_n-\frac{1}{n}J_n)$$ where $J_n$ is the $(n,n)$ matrix containing only ones. Finally $$(X_1,\ldots,X_k)|Y=0\sim N(0, I_k-\frac{1}{n}J_k)$$
Another fact: If $X\sim N(0,\Sigma)$ then $BX\sim N(0, B\Sigma B^T)$ (compute the Laplace transforms to see this). Therefore with $B=B_k$ then $$X_1+\cdots+X_k|Y=0\sim N(0, k-\frac{k^2}{n}).$$ Computing $\Pr(|X_1+\cdots+X_k|\leq \epsilon|Y=0)$ is now standard.