If $p=1$, $N=1$ and $a_1=(1/2,2/3,3/4,4/5,\ldots)$, the infimum equals 1 and is not achieved on a finitely supported vector (moreover, it is not achieved at all).

If $p=1$, $N=1$ and $a_1=(1/2,2/3,3/4,4/5,\ldots)$, the infimum equals 1 and is not achieved on a finitely supported vector (moreover, it is not achieved at all).

However if $0<p<1$ and the minimizer $x$ exists, it must have finite support (namely, of size at most $N$). To prove this, assume the contrary. Without loss of generality $x_1,\dots,x_{N+1}$ are positive. Choose a non-zero vector $b=(b_1,\dots,b_{N+1},0,0,\dots)$ orthogonal to all $a_i$'s. Choose small $t$ so that $x_i-t|b_i|>0>0$ for all $i=1,\dots,N+1$. Then by concavity of the function $x^p$ we have $$x_1^p+\ldots+x_{N+1}^p> \frac12\left((x_1+tb_1)^p+\ldots+(x_{N+1}+tb_{N+1})^p+\\+(x_1-tb_1)^p+\ldots+(x_{N+1}-tb_{N+1})^p\right).$$ Therefore one of the vectors $x\pm tb$ has smaller $p$-norm than $x$.

For $p>1$ the claim is completely false. Say, if $p=2$, $N=1$, the minimum is achieved on the vector proportional to $a_1$.

If $p=1$ and $a_n=1-1/n$, the infimum equals 1 and is not achieved on a finitely supported vector (moreover, it is not achieved at all).

If $p=1$, $N=1$ and $a_1=(1/2,2/3,3/4,4/5,\ldots)$, the infimum equals 1 and is not achieved on a finitely supported vector (moreover, it is not achieved at all).