Nice question! The answer is affirmative. If $\sigma(x)$ denotes the sum of primes up to $\sqrt{x}$, then it suffices to show that $\pi(x)-\sigma(x)$ changes sign infinitely often, because
$$\pi(x)-\sigma(x)<0<\pi(x+1)-\sigma(x+1)$$
never holds. I thank Juan Moreno and Will Sawin for this simple but crucial observation. My streamlined argument below owes to reuns' ideas as well.

Let us introduce the notation
$$\pi(x)=\mathrm{li}(x)-\mathrm{li}(1)+\rho(x),$$
then we get
\begin{align*}\sigma(x)
&=\int_1^\sqrt{x}t\,d\pi(t)\\
&=\int_1^\sqrt{x}\frac{t\,dt}{\log t}+\int_1^\sqrt{x} t\ d\rho(t)\\
&=\int_1^x\frac{du}{\log u}+\int_1^\sqrt{x} t\ d\rho(t)\\
&=\mathrm{li}(x)-\mathrm{li}(1)+\int_1^\sqrt{x} t\ d\rho(t).
\end{align*}
So the difference $\pi(x)-\sigma(x)$ can be directly estimated by the error term in the prime number theorem. We can analyze this difference further by considering the Mellin transform
$$\int_1^\infty x^{-s}\ d(\pi(x)-\sigma(x))=\int_1^\infty(x^{-s}-x^{1-2s})\ d\rho(x).$$
We shall only need to look at the left-hand side. Integrating by parts, we see that it is holomorphic in a region containing the half-plane $\Re(s)>1$ and the half-line $s>2/3$. In fact, in this region, the left-hand side equals
$$\log\zeta(s)-\log\zeta(2s-1)+\frac{1}{2}\log\zeta(4s-2)+f(s),\tag{$\ast$}$$
where $f(s)$ is holomorphic for $\Re(s)>2/3$.
As in the proof of Theorem 15.2 in Montgomery-Vaughan: Multiplicative number theory I, this eventually yields that
$$\pi(x)-\sigma(x)=\Omega_\pm(x^c)\quad\text{for any}\quad c<3/4.$$
Indeed, let us assume that this bound fails for some $c<3/4$. Without loss of generality, $c>2/3$. Then, by Landau's lemma (which is Lemma 15.1 in the same book), $(\ast)$ is holomorphic in the half-plane $\Re(s)>c$. However, this is easily seen to be false, and we are done.

in their capacity as mods,here. More specific to the current issue, though, the system expects people to edit their questions into a better shape, rather than self-delete and start again. If people see deletions and re-asking of questions, then it raises concerns, because this is a strategy that various bad actors have used in the past. $\endgroup$14more comments