Given an integer partition $\lambda$, introduce the following quantities:
\begin{align*}
c(\lambda)&=\sum_{i\geq1}\left\lceil\frac{\lambda_i}2\right\rceil, \qquad c_o(\lambda)=\sum_{i\geq1}\left\lceil\frac{\lambda_{2i-1}}2\right\rceil, \qquad
c_e(\lambda)=\sum_{i\geq1}\left\lceil\frac{\lambda_{2i}}2\right\rceil, \\
f(\lambda)&=\sum_{i\geq1}\left\lfloor\frac{\lambda_i}2\right\rfloor, \qquad f_o(\lambda)=\sum_{i\geq1}\left\lfloor\frac{\lambda_{2i-1}}2\right\rfloor, \qquad
f_e(\lambda)=\sum_{i\geq1}\left\lfloor\frac{\lambda_{2i}}2\right\rfloor, \\
a_o(\lambda)&=\sum_{i\geq1}\lambda_{2i-1}, \qquad
a_e(\lambda)=\sum_{i\geq1}\lambda_{2i}.
\end{align*}
With these in mind, we have the following identities:
QUESTIONS. All sums run through all partitions of the integer $n$. Are these true?
(1) $\sum_{\lambda\vdash n}f_o(\lambda)=\sum_{\lambda\vdash n}c_e(\lambda)$;
(2) $\sum_{\lambda\vdash n}a_o(\lambda)=\sum_{\lambda\vdash n}c(\lambda)$;
(3) $\sum_{\lambda\vdash n}a_e(\lambda)=\sum_{\lambda\vdash n}f(\lambda)$;
(4) $\sum_{\lambda\vdash n}(c_o(\lambda)-f_e(\lambda))
=\sum_{\lambda\vdash n}(a_o(\lambda)-a_e(\lambda))$.
Example. These are $\{(5), (4,1), (3,2), (3,1,1), (2,2,1), (2,1,1,1), (1,1,1,1,1)\}$ the partitions of $n=5$. The relevant vectors for the identities in question are:
$f_o$ in $\{(2), (2), (1), (1,0), (1,0), (1,0), (0,0,0)\}$
$c_e$ in $\{(0), (1), (1), (1), (1), (1,1), (1,1)\}$
$a_o$ in $\{(5), (4), (3), (4), (3), (3), (3)\}$
$c$ in $\{(3), (3), (3), (4), (3), (4), (5)\}$
$a_e$ in $\{(0), (1), (2), (1), (2), (2), (2)\}$
$f$ in $\{(2), (2), (2), (1), (2), (1), (0)\}$
$c_o$ in $\{(3), (2), (2), (3), (2), (2), (3)\}$
$f_e$ in $\{(0), (0), (1), (0), (1), (0), (0)\}$.
Remark. I can provide a generating function proof of (4), maybe others too. So, I only wish to see a combinatorial justification for each assertion.
Best Answer
It's easy to check that all four identities are true when the sum is restricted to two terms indexed by a partition $\lambda$ and its conjugate $\lambda'$. (If $\lambda=\lambda'$ then the two terms are equal, so we only need one term.) From this (1)-(4) follow by grouping the terms in conjugate pairs (or just one term when $\lambda=\lambda'$).
Here is an illustrative example for the pairing of $\lambda$ and its conjugate $\lambda'$. Just for identity (1).
Let $\lambda=(5,4,2)$ so that $\lambda'=(3,3,2,2,1)$. Fill in the odd-indexed rows by alternating $a$'s and $b$'s; fill in the even-indexed rows by alternating $c$'s and $d$'s (see this paper, pages 1-2). In the present case, these look like:
$$\lambda=\begin{array} 1 a&{\mathbf{\color{blue}b}}&a&{\mathbf{\color{blue}b}}&a \\ {\mathbf{\color{red}c}}&d&{\mathbf{\color{red}c}}&d \\ a&{\mathbf{\color{blue}b}}\end{array} \qquad \lambda'=\begin{array} 1 a&{\mathbf{\color{blue}b}}&a \\ {\mathbf{\color{red}c}}&d&{\mathbf{\color{red}c}} \\ a&{\mathbf{\color{blue}b}} \\ {\mathbf{\color{red}c}}&d \\ a \end{array} $$ We notice that $f_o(\mu)=\#$ of ${\mathbf{\color{blue}b}}$'s while $c_e(\mu)=\#$ of ${\mathbf{\color{red}c}}$'s in the respective diagrams. Therefore, $$f_o(\lambda)+f_o(\lambda')=3+2=2+3=c_e(\lambda)+c_e(\lambda').$$