In an earlier post, user Robert Z proved $$\sum_{k=0}^{n}p_{2k+1} \sim 2 n^2\ln n$$ $$\sum_{k=0}^{n}p_{2k} \sim 2 n^2\ln n$$
Yet clearly (the $2k$ is arbitrary here, replace it with $k$ if you feel so inclined)
$$\sum_{k=0}^{n}p_{2k+1}-p_{2k} \nsim 0$$
Although by following Robert's method, I am unable to show that the above difference is asymptotic to anything useful. Can a meaningful $f(n)$ asymptotic to the above sum of differences be determined? Meaningful, as in non-constant.
I believe it's reasonable to assume the prime gaps from consecutive even or odd primes should not, on average, be very much different compared to each other, so summing them should give approximately the same values. As such, you get
$$\sum_{k=1}^{n}p_{2k+2}-p_{2k+1} \approx \sum_{k=1}^{n}p_{2k+1}-p_{2k} \approx \sum_{k=1}^{n}p_{2k}-p_{2k-1} \tag{1}\label{eq1A}$$
Actually, as the prime gaps tend to, on average, be growing for larger primes, especially for large enough $n$, you can often replace the $\approx$ with $\ge$ in \eqref{eq1A} to have upper & lower bounds of the middle expression value. Next, let
$$m = \sum_{k=1}^{n}p_{2k+1}-p_{2k} \tag{2}\label{eq2A}$$
Then \eqref{eq1A} becomes
$$p_{2n+2} - m - p_2 \approx m \approx p_{2n+1} - m - p_1 \tag{3}\label{eq3A}$$
Add $m$ to all $3$ parts and divide by $2$ to get
$$\frac{p_{2n+2} - p_2}{2} \approx m \approx \frac{p_{2n+1} - p_1}{2} \tag{4}\label{eq4A}$$
Thus, by the Prime Number Theorem approximation, you have $p_{2n} \sim (2n)\log(2n)$, so you then get
$$\sum_{k=0}^{n}p_{2k+1}-p_{2k} \sim \frac{(2n)\log(2n)}{2} \sim n\log(n) \tag{5}\label{eq5A}$$
This matches what Lord Shark the Unknown's question comment says the asymptotic value should be.
OP requests: Meaningful, as in non-constant.
Let's do it:
$$ p_1-p_0=1\qquad\mbox{and}\qquad \forall_ {n\ge 1}\,\, p_{2\cdot n+1}-p_{2\cdot n}\ge 2 $$ hence:
$$ \sum_{k=0}^n\,\,(p_{2\cdot n+1}-p_{2\cdot n})\,\,\ge \,\,2\cdot n-1 $$
This has fulfilled the OP's request.
It seems not easy to improve upon the above inequality meaningfully by elementary means. The following equation is a symptom:
$$ p_3-p_2\,=\,p_5-p_4\,=\,p_7-p_6\,=\,2 $$
