Let $A$ be the set of numbers $x$ whose binary expansion has 0's and 1's in equal proportion, i.e., if $w(x,n)$ is the number of 1's in the first $n$ places after the binary-point (I can't say "decimal point" in binary notation) in the binary expansion of $x$, then $w(x,n)/n\to 1/2$ as $n\to\infty$. By the strong law of large numbers, the complement $U$ of $A$ has measure zero. I claim that this $U$ has the property asked for in the question. To prove it, let arbitrary $p_1,\dots,p_n$ be given; I'll find an $a$ such that all the $p_i+a$ are in $U$, i.e., none of them has 0's and 1's in equal proportion. I'll find $a$ by gradually building its binary expansion. Begin by writing "0."; now I have to fill in what comes after that binary-point. Fill in a large but finite number, say $N_1$, of places in $a$ so as to make the first $N_1$ places in $p_1+a$ all zero; so $w(p_1+a,N_1)=0$. (Of course this gives me no control over $w(p_i+a,N_1)$ for $i\neq 1$.) Next, fill in a much larger number, say $N_2$, of the subsequent places of $a$ so as to make all those places in $p_2+a$ zero. As a result, $w(p_2+a,N_1+N_2)$ is very small; the point is that the $N_2$ zeros outweigh whatever happened in the first $N_1$ places of $p_2+a$ because $N_2$ is way bigger than $N_1$. (Of course, I now have no control over $w(p_i+a,N_1+N_2)$ for $i\neq2$.) Continue in the same way, choosing an even larger number $N_3$ of the subsequent places in $a$ so as to make $w(p_3+a,N_1+N_2+N_3)$ very small. After $n$ steps like this, culminating in $w(p_n+a,N_1+\dots+N_n)$ being very small, cycle back to work with $p_1$, making $w(p_1+a,N_1+\dots+N_n+N_{n+1})$ very small. Keep cycling around forever. For each $p_i$, you'll get infinitely many $M$'s (sums of appropriate numbers of $N_k$'s) for which $w(p_i+a,M)$ is very small. Therefore $p_i+a\in U$, as claimed. (Confession: I've been sloppy about any carrying that would be involved when the sums $p_i+a$ are done in binary notation; un-slopping the argument is left as an exercise.)