Here is a partial idea ~~ where the last step needs a bit more justification ~~ which doesn't work as well as I had hoped, but may yet have promise.

As noted, this is a question about adding real numbers, no specialized measure theory is involved although the language is convenient. ~~ We will ~~ had hoped to define $2^n$ values $t_I$, one for each $I \subseteq [n]=\lbrace1,2,\cdots,n\rbrace $ so that if they are all non-negative, then the specified values can be achieved, otherwise they can not.

Suppose that we have a set $U$ of size (or measure) $m_{\emptyset}=|U|$ along with $n$ subsets $X_i$ for $i \in [n]$ each with complement $\overline{X_i}.$ For each $I \subseteq [n],$ let $m_I=|\cap_{i \in I}X_i|$ while $t_I=|(\cap_{i \in I}X_i) \cap (\cap_{j\notin I}\overline{X_j})|.$ As in the problem we can abbreviate $m_i=m_{i,i}$ for $m_{\lbrace i \rbrace}=|X_i|$ and $m_{i,j}$ for $m_{\lbrace i,j\rbrace}=|X_i\cap X_j|.$

If we know either set of values we can uniquely find the others: $$m_I=\sum_{I \subseteq J \subseteq [n]}t_J\hspace{0.1in} \text{ while } t_I=\sum_{I \subseteq J \subseteq [n]}(-1)^{|J|-|I|}m_J$$

If the $m_I$ are given, then the $t_I$ will be determined over the reals, but we want non-negative values. In the given problem we have only $n+\binom{n}{2}$ of the $m$ values, perhaps along with $m_{\emptyset}=1.$

So we first define the rest of the $m_I$ by $m_I=\min_{i,j \in I}m_{i,j}$ then solve for the $t_I$ and check that they are non-negative. I think that these choices will make $t_{[n]}$ as large as possible and thought that they would give all the $t_I$ the best chance to be non-negative consistent with the given information.

**BUT** now I notice problems with the simple case of asking for the sides of a triangle: $|U|=3$ $|X_1|=|X_2|=|X_3|=2$ and $|X_1 \cap X_2|=|X_1 \cap X_3|=|X_2 \cap X_3|=1$ If we do add the final condition $m_{\lbrace 1,2,3 \rbrace}=|X_1 \cap X_2 \cap X_3|=0$ then we do get the desired solution $t_{\{1,2\}}=t_{\{1,3\}}=t_{\{2,3\}}=1$ with the rest of the $t_I=0.$ HOWEVER if we make my suggested choice of $m_{\lbrace 1,2,3 \rbrace}=|X_1 \cap X_2 \cap X_3|=1$ then we do get $t_{\{1,2,3\}}=1$ as large as possible but then $t_{\{1,2\}}=t_{\{1,3\}}=t_{\{2,3\}}=0$ making $t_{\{1\}}=t_{\{2\}}=t_{\{3\}}=1$ and finally $t_{}=-1$

Think of the values $t_I$ as weights to be assigned to the regions of an ideal Venn diagram for $n$ sets. The conditions on the $m_i$ and $m_i,j$ (along perhaps with $m_{\emptyset}=1$) give $\frac{n^2+n}2$ (or else $\frac{n^2+n}2+1$ ) equations in $2^n$ variables $t_I$ along with the side condition that all the $t_I \ge 0.$ This is a linear programming problem. Perhaps there is a way to start with my assignmet and then adjust the values until success or proved failure, but I am not sure.

**later** The convex hull might be unworkable once the number of dimensions is $43$ or $43+\binom{43}{2}$ or $2^{43}$, but maybe not. I found the comments on integral problems pretty convincing and asked this question. However I subsequently answered it and realized that, while there is no projective plane of order $6$, We can achieve $m_{\emptyset}=43, m_i=7$ for $1 \le i \le 43$ and all $m_{i,j}=1$ by setting $|t_I|=0$ except that $|t_J|=1/\binom{41}{5}$ for all $\binom{43}{7}$ choices of $J$ with $|J|=7.$ So now I am back to thinking that there might be a simple algorithm which achieves the constraints with least $L_1$ error (so $0$ if possible.)