The answer is "no". Here's a proof:
The main idea is to consider the two sums as approximations of the same integral over $1/x$ with the rectangle rule and then show that the approximation errors can't be equal.
It will be more convenient to work with the centre of each sum rather than the beginning. So let $k_i$ be the number of terms in the $i$th sum (as in the question), let $n_i=m_i + (k_i-1)/2$ be the central denominator (which may or may not actually occur as the denominator of a term in the sum depending on whether it's an integer or a half-integer), and let $\sigma_i = \sigma(m_i,k_i)$. Without loss of generality, assume $n_2 > n_1$, and as Henry observed, we can also assume that the sums don't overlap.
Now $\sigma_i$ is roughly $k_i/n_i$, so the quantity $\Delta:=k_1n_2-k_2n_1$ would need to be small in order for the sums to be equal. This quantity is either an integer or a half-integer. Note that if $k_1$, $k_2$ and $n_2$ are fixed, then $\sigma_2$ is fixed, $n_1$ depends monotonically on $\Delta$, and $\sigma_1$ in turn depends monotonically on $n_1$, so the difference between the two sums can change sign at most once as $\Delta$ changes. Thus, if we can show that the difference has opposite signs for $\Delta = 0$ and $\Delta = 1/2$, it will follow that it cannot be zero for any possible value of $\Delta$.
Now consider $1/l$ as the first term in the following expansion:
$$\int_{l-\frac{1}{2}}^{l+\frac{1}{2}}\frac{\mathrm{d}x}{x} = \int_{-\frac{1}{2}}^{+\frac{1}{2}}\frac{\mathrm{d}x}{l-x}=\int_{-\frac{1}{2}}^{+\frac{1}{2}}\frac{1}{l}\sum_{i=0}^\infty\left(\frac{x}{l}\right)^i\mathrm{d}x=2\sum_{j=0}^\infty\frac{\left(2l\right)^{-(2j+1)}}{2j+1}\;.$$
Then $\sigma_1$ is the first term of the following expansion:
$$\int_{n_1-k_1/2}^{n_1+k_1/2}\frac{\mathrm{d}x}{x}=\sum_{l=n_1-(k_1-1)/2}^{n_1+(k_1-1)/2}\int_{l-\frac{1}{2}}^{l+\frac{1}{2}}\frac{\mathrm{d}x}{x}=2\sum_{j=0}^\infty\frac{1}{2j+1}\sum_{l=n_1-(k_1-1)/2}^{n_1+(k_1-1)/2}\left(2l\right)^{-(2j+1)}\;.$$
Now consider first the case $\Delta=0$. Then $n_2=\frac{k_2}{k_1}n_1$, so by substituting $u=\frac{k_2}{k_1}x$ we can write the two sums as approximations of the same integral:
$$\int_{n_1-k_1/2}^{n_1+k_1/2}\frac{\mathrm{d}x}{x}=\int_{n_2-k_2/2}^{n_2+k_2/2}\frac{\mathrm{d}u}{u}=2\sum_{j=0}^\infty\frac{1}{2j+1}\sum_{l=n_2-(k_2-1)/2}^{n_2+(k_2-1)/2}\left(2l\right)^{-(2j+1)}\;,$$
where the $j=0$ term of this expansion is now $\sigma_2$. Since these are two expansions of the same integral, to show $\sigma_1<\sigma_2$ it suffices to show that for each $j>0$ the term in the first expansion is greater than the term in the second expansion. (This is plausible, since in going from the first approximation of the integral to the second, we've increased the number of intervals by a factor $k_2/k_1$ but decreased the $j$th error term in each interval roughly by a factor $(n_1/n_2)^{2j+1}$, which is $(k_1/k_2)^{2j+1}$ since $\Delta=0$.) In fact, we only need to treat the $j=1$ case; the inequalities for higher $j$ then follow since all additional factors of $l^{-2}$ are greater in the first expansion than in the second expansion, since we assumed that the two sums of reciprocals don't overlap. (We could have avoided the terms with higher $j$ altogether by writing the error term as a derivative at an intermediate value, but that would have introduced cumbersome shifts of $1/2$ to account for the intermediate values.)
For $j=1$, we use the convexity of $l^{-3}$ in both directions, decreasing the greater one of the sums by collapsing it onto its center:
$$\sum_{l=n_1-(k_1-1)/2}^{n_1+(k_1-1)/2}l^{-3}\ge\sum_{l=n_1-(k_1-1)/2}^{n_1+(k_1-1)/2}n_1^{-3}=\frac{k_1}{n_1^{3}}$$
and increasing the lesser one by smearing it out over an integral:
$$\sum_{l=n_2-(k_2-1)/2}^{n_2+(k_2-1)/2}l^{-3}<\int_{n_2-k_2/2}^{n_2+k_2/2}l^{-3}\mathrm{d}l=\frac{(n_2-k_2/2)^{-2}-(n_2+k_2/2)^{-2}}{2}=$$
$$=\frac{n_2k_2}{(n_2-k_2/2)^2(n_2+k_2/2)^2}<\frac{n_2k_2}{n_1^2n_2^2}=\frac{k_1}{n_1^{3}}\;.$$
This establishes that $\sigma_1<\sigma_2$ when $\Delta=0$. Now consider $\Delta \neq 0$. Then $\frac{k_1}{k_2}n_2 = n_1+\frac{\Delta}{k_2}$, and we need to shift the integrand to make the integral limits match:
$$\int_{n_2-k_2/2}^{n_2+k_2/2}\frac{\mathrm{d}x}{x}=\int_{n_1-k_1/2+\Delta/k_2}^{n_1+k_1/2+\Delta/k_2}\frac{\mathrm{d}u}{u}=\int_{n_1-k_1/2}^{n_1+k_1/2}\frac{\mathrm{d}t}{t+\Delta/k_2}\;.$$
Since all the error terms in the expansions for the integrals have the same sign, their difference is less than the larger of the two, which, as we showed above, is the one for $\sigma_1$. Thus, to show that the difference changes sign when $\Delta$ goes from $0$ to $1/2$, it suffices to show that the change due to the shift in the integrand is larger than the approximation error in the expansion for $\sigma_1$. To do this, we can expand the shifted integral in the same form as the error expansion, again using convexity to bound each integral by the central value of the integrand:
$$\int_{n_1-k_1/2}^{n_1+k_1/2}\frac{\mathrm{d}t}{t+\Delta/k_2}=\sum_{l=n_1-(k_1-1)/2}^{n_1+(k_1-1)/2}\int_{l-\frac{1}{2}}^{l+\frac{1}{2}}\frac{\mathrm{d}t}{t+\Delta/k_2}>\sum_{l=n_1-(k_1-1)/2}^{n_1+(k_1-1)/2}\frac{1}{l+\Delta/k_2}\;.$$
To show that for each $l$ the summand differs from $1/l$ by more than the corresponding approximation error, we estimate the latter so that we can sum the series over $j$:
$$2\sum_{j=0}^\infty\frac{1}{2j+1}\left(2l\right)^{-(2j+1)}<
\frac{2}{3}\sum_{j=0}^\infty\left(2l\right)^{-(2j+1)}=\frac{2}{3}\left(2l\right)^{-3}\frac{1}{1-1/(2l)}=\frac{2}{3}\frac{1}{(2l)^2}\frac{1}{2l-1}\;.$$
On the other hand, the difference from the shift in the reciprocals is
$$\frac{1}{l}-\frac{1}{l+\Delta/k_2}=\frac{\Delta/k_2}{l(l+\Delta/k_2)}\;.$$
Now we can estimate the quotient of these two values (using $\Delta=1/2$):
$$\frac{2}{3}\frac{1}{(2l)^2}\frac{1}{2l-1}\frac{l(l+\Delta/k_2)}{\Delta/k_2}<
\frac{1}{12}\frac{1}{l}\frac{1}{l-1/2}\frac{l+1/2}{\Delta/k_2}<
\frac{k_2}{6(l-1)}\;.$$
Thus, the change due to the shift in the integrand is larger than the approximation error provided $k_2<6(l-1)$. This we can derive by considering the powers of $2$ in the denominators of the two sums.
The highest power of $2$ that divides one of the denominators in the sum cannot be cancelled and hence divides the reduced denominator of the sum. This is because between any two numbers containing the same number of factors of $2$, there is one containing at least one more factor of $2$, and hence each sum contains a unique denominator $d$ with the highest power of $2$ in that sum. If we add all the other reciprocals and reduce them to a common denominator, that denominator will necessarily contain fewer powers of $2$ than $d$, and hence these powers cannot cancel if we then add $1/d$. Thus, the two sums can only be equal if the denominator with the most factors of $2$ has the same number of factors of $2$ in both sums.
For this to be the case, $k_2$ must be at least $1$ less than twice the largest denominator in $\sigma_1$; otherwise any interval of $k_2$ numbers would necessarily contain a number with one factor of $2$ more than the highest power of $2$ in the denominators of $\sigma_1$. Thus we have
$$k_2 \le 2\left(n_1+\frac{k_1-1}{2}\right)-1=2n_1+k_1-2\;.$$
Since the sums don't overlap, we also have
$$n_1+\frac{k_1-1}{2}\le n_2-\frac{k_2-1}{2}-1\;,$$
and hence
$$\frac{k_2}{2}\le \left(n_1+\frac{k_1-1}{2}\right)-\frac{1}{2} \le n_2-\frac{k_2-1}{2}-1-\frac{1}{2}\;,$$
$$k_2\le n_2-1\;.$$
With $\Delta=k_1n_2-k_2n_1= 1/2$, this yields
$$k_1=\frac{k_2}{n_2}n_1+\frac{\Delta}{n_2}\le n_1-\frac{n_1}{n_2}+\frac{\Delta}{n_2}<n_1\;,$$
and hence $k_1\le n_1 - 1/2$, since $k_1$ is an integer and $n_1$ an integer or a half-integer. Thus
$$k_2\le2n_1+k_1-2<3n_1-2<6(\frac{n_1}{2}-\frac{1}{4})\le 6\left(n_1-\frac{k_1-1}{2}-1\right)\le 6(l-1)\;.$$
This completes the proof.
P.S.: In case you're surprised that we get exactly the factor of $6$ that we need from considering the powers of $2$: I am, too :-) If there's any deeper reason behind this, I have no idea what it might be. It's possible to derive a slightly tighter bound using Jitsuro Nagura's improvement on Bertrand's postulate, since, as Henry observed, any prime in the second sum that is greater than half its endpoint is enough to make the reduced denominators differ. But I think such improvements in bounding $k_2$ with respect to $n_2$ can't make the factor in the final inequality lower than $4$, since they only improve the estimate of $k_1$ with respect to $n_1$, but not that of $k_2$ with respect to $n_1$ from the powers of $2$, which is $k_2 < 2 n_1$ even for $k_1=1$.
Best Answer
We can use the formula $$ \frac1n = \frac1{n+1} + \frac1{n(n+1)} $$ to obtain such a sum. Start with $$ 1 = \frac12 + \frac14 + \frac18 + \frac18 $$ Now, for $n=8$ we can rewrite the last term as $$ \frac18=\frac19+\frac{1}{72} $$ Since $72+1$ is a prime, we have to iterate this, i.e., $$ \frac18=\frac19+\frac{1}{73}+\frac{1}{5256}. $$ Similarly $$ \frac12=\frac13+\frac17+\frac{1}{43}+\frac{1}{1806} $$ and $$ \frac14=\frac15+\frac{1}{20}. $$ Together we obtain $$ 1=\left(\frac13+\frac17+\frac{1}{43}+\frac{1}{1806}\right)+\left(\frac15+\frac{1}{20}\right)+\frac18+\left(\frac19+\frac{1}{73}+\frac{1}{5256}\right). $$ All denominators plus one are not prime numbers.