3. Find all positive integers $n$ such that any 35-element subset of the set
$$
M=\{1,2, \cdots, 50\}
$$

contains at least two distinct elements $a, b$ satisfying $a+b=n$ or $a-b=n$.
3. Let $A=\{1,2, \cdots, 35\}$.

Then for any $a, b \in A$, we have
$$
a-b, a+b \leqslant 34+35=69.
$$

On the other hand, let $A=\left\{a_{1}, a_{2}, \cdots, a_{35}\right\}$, and
$$
a_{1}<a_{2}<\cdots<a_{35}.
$$
(1) When $1 \leqslant n \leqslant 19$, consider
$$
\begin{array}{l}
1 \leqslant a_{1}<a_{2}<\cdots<a_{35} \leqslant 50, \\
1 \leqslant a_{1}+n<a_{2}+n<\cdots<a_{35}+n \\
\leqslant 50+19=69.
\end{array}
$$

By the pigeonhole principle, there exist $1 \leqslant i<j \leqslant 35$, such that
$$
a_{i}+n=a_{j} \Rightarrow a_{i}-a_{j}=n.
$$
(2) When $51 \leqslant n \leqslant 69$, consider
$$
\begin{array}{l}
1 \leqslant a_{1}<a_{2}<\cdots<a_{35} \leqslant 50, \\
1 \leqslant n-a_{35}<n-a_{34}<\cdots<n-a_{1} \leqslant 68.
\end{array}
$$

By the pigeonhole principle, there exist $1 \leqslant i<j \leqslant 35$, such that
$$
n-a_{i}=a_{j} \Rightarrow a_{i}+a_{j}=n.
$$
(3) When $20 \leqslant n \leqslant 24$, since
$$
50-2 n \leqslant 50-40=10,
$$

then at least 25 of $a_{1}, a_{2}, \cdots, a_{35}$ belong to $[1,2 n]$.
$$
\text{Also, }\{1, n+1\},\{2, n+2\}, \cdots,\{n, 2 n\} \text{ are at most 24, so there exists}
$$

$$
\left(a_{i}, a_{j}\right)=(t, n+t),
$$

i.e., $a_{j}-a_{i}=n$.
(4) When $25 \leqslant n \leqslant 34$, note that $\{1, n+1\}$, $\{2, n+2\}, \cdots,\{n, 2 n\}$ are at most 34.
By the pigeonhole principle, there exist
$$
\left(a_{i}, a_{j}\right)=(t, n+t) \Rightarrow a_{j}-a_{i}=n.
$$
(5) When $35 \leqslant n \leqslant 50$, if $n=2 k+1$, then
$$
\begin{array}{l}
\{1,2 k\},\{2,2 k-1\}, \cdots,\{k, k+1\}, \\
\{2 k+1\}, \cdots,\{50\}
\end{array}
$$

are a total of $50-k$.
Hence, there exist $1 \leqslant i<j \leqslant 35$, such that
$$
a_{i}+a_{j}=2 k+1=n.
$$

If $n=2 k$, then
$$
\begin{array}{l}
\{1,2 k-1\},\{2,2 k-2\}, \cdots,\{k-1, k+1\}, \\
\{k\},\{2 k\},\{2 k+1\}, \cdots,\{50\}
\end{array}
$$

are a total of $51-k$.
Hence, there exist $1 \leqslant i<j \leqslant 35$, such that $a_{i}+a_{j}=2 k=n$. In summary, the required $n$ is the integers from $1$ to $69$.