Three same-color integers

Pairs. $n$ pairs, each with difference $n$.

Pigeonhole. Color each pair with $2$ colors. To avoid a same-color pair, we'd need $n\le 2$ (only $2$ colors). For $n\ge3$, some pair has the same color.

Actually, the simple pigeonhole argument doesn't directly apply: $2$ colors across $n$ pairs doesn't force a same-color pair, since each pair can be different colors.

Better approach. Suppose no pair of same-color integers differs by $n$. Then for each pair $(i,i+n)$, the two colors differ. This forces the coloring pattern: color $i$ and color $i+n$ are opposite for each $i\in\{1,\ldots,n\}$. But then color $i$ and $i+2n$? We only have $2n$ integers.

Conclusion. The statement is true by a parity argument for $2$ colors: color classes must pair up to have opposite colors, which is possible. Actually the statement is false: color the first $n$ red and the last $n$ blue; then each pair $(i,i+n)$ has one red, one blue — no same-color pair with difference $n$.

Correction. The problem is false as stated. The correct Ramsey-type statement needs different parameters.