Consider $n$ independent one-dimensional Wiener processes $(W_i)_{1\leqslant i\leqslant n}$.
Is there with probability $1$ some time $t\in[0,1]$ such that $W_i(t)>0$ for every $1\leqslant i\leqslant n$?
For $n=1$ this is obvious, but I don't see how to show this for $n\geqslant2$.
Yes, for every $n$. Here is a proof.
The distribution of each $W_i$ is invariant by the transformation $t\mapsto tW_i(1/t)$, hence the probability to be computed is also the probability that there exists some $t\gt1$ such that $W_i(t)\gt0$ simultaneously for every ${1\leqslant i\leqslant n}$. We now define iteratively an increasing sequence of random times as follows.
First, note that, for any Brownian motion $W$, $P(W(t)\gt0\mid W(0)=-1)\to\frac12$ when $t\to\infty$ hence there exists some deterministic and finite $\theta$ such that $P(W(\theta)\gt0\mid W(0)=-1)\geqslant\frac13$. Then, by scaling, for every $x$, $$ P(W(x^2\theta)\gt0\mid W(0)=x)\geqslant\tfrac13. $$ Let $T_0=1$. For every $k\geqslant0$, knowing $T_k$, let $$ T_{k+1}=T_k+\theta\cdot\max\limits_{1\leqslant i\leqslant n}W_i(T_k)^2. $$ Then, almost surely, for every ${1\leqslant i\leqslant n}$, $$ P(W_i(T_{k+1})\gt0\mid \mathcal F^{W_i}_{T_k})\geqslant\tfrac13. $$ Since the Brownian motions $W_i$ are independent, this implies that $P[A_{k+1}\mid\mathcal G_k]\geqslant3^{-n}$, where $$ A_k=[\forall 1\leqslant i\leqslant n,\,W_i(T_{k})\gt0],\qquad\mathcal G_k=\sigma(A_\ell;\,\ell\leqslant k). $$ In particular, $P[A_{k+1}^c\mid A_1^c\cap\cdots\cap A_k^c]\leqslant1-3^{-n}$ hence the probability that no event $A_\ell$ with $\ell\leqslant k$ holds is at most $(1-3^{-n})^k$, which goes to zero when $k\to\infty$. Conversely, the event $A_k$ holds for at least some $k$, almost surely, in particular, almost surely there exists some $t$ such that $W_i(t)\gt0$ simultaneously for every ${1\leqslant i\leqslant n}$. QED.
The same approach shows that, for every level $h$, almost surely there exists some $t$ such that $W_i(t)\gt h$ simultaneously for every ${1\leqslant i\leqslant n}$.