During my research, I checked using Maple that we have numerically for $ 2 \leq n \leq 20,$ $ \forall t $ integer satisfying $ 0 \leq t \leq n-1$


$$ \displaystyle \frac{1}{2} \displaystyle \sum\_{p=1}^{n-t} C\_{n}^{p+t} C\_{n+p+t}^{n} \displaystyle \frac{(-1)^{p}}{p}= \displaystyle C\_{n+t}^n\displaystyle \sum\_{p=1}^{n-t} C\_{n}^{p+t} \displaystyle \frac{(-1)^{p}}{p}$$


I think that this true for all integer $n \geq 2$ and forall $t$ integer satisfying $\forall 0 \leq t \leq n-1$ and i need a proof... any help?


Is it known result? ps: here $ \forall m,k$ integer with $0 \leq k \leq m, C\_m^k=\displaystyle \frac{m!}{k!(m-k)!}$



About binomial identity
We have $$C\_{n}^{p+t}C\_{n+p+t}^n=[y^{n-t}x^tz^t](x+y+z)^{n+t}\cdot \left(\frac{(x+y+z)y}{xz}\right)^p,$$
where $[M]f$ is a coefficient of monomial $M$ in the polynomial or series $f$. Multiplying by $(-1)^p/p$ and summing by all $p=1,2,\ldots$ we get
$$\sum\_{p=1}^\infty (-1)^p p^{-1}C\_{n}^{p+t}C\_{n+p+t}^n=[y^{n-t}x^tz^t] (x+y+z)^{n+t}\log\left(1+\frac{(x+y+z)y}{xz}\right).
$$
It remains to use the identity 
$$
\log\left(1+\frac{(x+y+z)y}{xz}\right)=\log\left(1+\frac{y}x\right)+\log\left(1+\frac{y}z\right)
$$
to get twice the right hand side of your formula:
$$
[y^{n-t}x^tz^t](x+y+z)^{n+t}\log\left(1+\frac{y}x\right)=C\_{n+t}^t
[y^{n-t}x^t](x+y)^{n}\log\left(1+\frac{y}x\right),
$$
and
$$
[y^{n-t}x^t](x+y)^{n}\log\left(1+\frac{y}x\right)=\sum\_{p=1}^{n-t}
\frac{(-1)^p}p [y^{n-t-p}x^{t+p}] (x+y)^{n}=
\sum\_{p=1}^{n-t} C\_{n}^{p+t} \displaystyle \frac{(-1)^{p}}{p}
$$


We should explain what is the ring the series above belong to, let it be the ring of power series in $x,z$ and $y/(xz)$.


