Are regular languages closed against an intersection that keeps words with the same number of ones?
Let 
$$
  L = \{ x \in L_1 \mid \text{there exists }y \in L_2 \text{ such that } |x|_1 = |y|_1\}
$$
Let $\pi: \Sigma^* \to a^*$ be the homomorphism defined by $\pi(u) = a^{|u|_1}$.
I claim that
$$
  L = L_1 \cap \pi^{-1}(\pi(L_2))
$$
Since regular languages are closed under homomorphisms, inverses of homomorphisms and intersection, it will follow that $L$ is regular.

Proof of the claim.
\begin{align}
  L_1 \cap \pi^{-1}(\pi(L_2)) &amp;= \{ x \in L_1 \mid \pi(x) \in  \pi(L_2) \} \\
  &amp;= \{ x \in L_1 \mid \text{there exists }y \in L_2 \text{ such that } \pi(x) = \pi(y)\}\\
  &amp;= \{ x \in L_1 \mid \text{there exists }y \in L_2 \text{ such that } |x|_1 = |y|_1\} \\
  &amp;= L
\end{align}