Let $(a_{n,k}: n,k \ge 1)$ be an infinite matrix of positive reals such that $\sum_{k\ge 1}a_{n,k}=1$ for all $n$.
Question. Is it true that there exist a free ultrafilter $\mathscr{F}$ on $\mathbf{N}$ and a real sequence $(x_n: n\ge 1)$ such that $$\mathscr{F}\text{-}\lim_n \sum_{k\ge 1}|a_{n,k}-x_n|=0?$$
Ps. We recall that, if $(y_n: n\ge 1)$ is a real sequence and $\ell \in \mathbf{R}$, then $\mathscr{F}\text{-}\lim_n y_n=\ell$ means that $\{n \in \mathbf{N}: |y_n-\ell| < \varepsilon\} \in \mathscr{F}$ for all $\varepsilon>0$.
Let $a_{n,k}=[n=k]$, where the righthand side is an Iverson bracket, and let $\mathscr{U}$ be a free ultrafilter on $\Bbb Z^+$. Let $\langle x_n:n\in\Bbb Z^+\rangle$ be a real sequence; then
$$|a_{n,k}-x_n|=\begin{cases} |1-x_n|,&\text{if }n=k\\ |x_n|,&\text{otherwise,} \end{cases}$$
so
$$\sum_{k\ge 1}|a_{n,k}-x_n|=|1-x_n|+\sum_{\substack{k\ge 1\\k\ne n}}|x_k|\,.$$
Suppose that $\mathscr{U}$-$\lim\sum_{k\ge 1}|a_{n,k}-x_n|=0$. Then there is a $U\in\mathscr{U}$ such that
$$|1-x_n|+\sum_{\substack{k\ge 1\\k\ne n}}|x_k|<\frac12$$
for all $n\in U$. Let $m,n\in U$ with $m\ne n$. Then $n\in U$ implies that $|1-x_n|<\frac12$, and $m\in U$ implies that $|x_n|<\frac12$. Clearly this is impossible. Thus, the conclusion is false for the infinite identity matrix.