For a given infinite set of primes, not too big, eg, satisfying LangTrotter conjecture, can we always find an E.C. with supersingular reduction (at least) at these primes? How about E.C. without CM?
No. In fact the set can be arbitrarily sparse, i.e. the $n$th prime in the set can be chosen to exceed $a_n$ for any sequence $\{a_n\}$. This is because the rationals are countable. Fix an enumeration $j_1,j_2,j_3,\ldots$ of $\bf Q$. For each $n$ let $p_n$ be the smallest prime such that $p_n > a_n$ and $j_n$ is not supersingular mod $p_n$ (this is possible because the density of ordinary primes is positive, namely $1/2$ if $j_n$ is one of the thirteen CM $j$invariants and $1$ otherwise). Then no rational number is the $j$invariant of an elliptic curve with supersingular reduction at every $p_n$.

$\begingroup$ What do you mean by "thirteen CM jinvariants"? I don't know this result. Could you explain it or give me some reference? $\endgroup$ – user42690 Nov 24 '13 at 7:11

3$\begingroup$ @user42690: What Noam Elkies is refering to is that there exixts exactly $13$ values of $j$ which are in $\mathbb{Q}$ and whose corresponding elleptic curves is CM. These are completely explicit and correspond to the $13$ orders in imaginary quadratic fields which are principal domains. $\endgroup$ – Filippo Alberto Edoardo Nov 24 '13 at 8:46

$\begingroup$ @user42690: Look at Noam's answer there mathoverflow.net/questions/149702/… $\endgroup$ – ACL Nov 24 '13 at 9:57