Is SHA1 secure with such many inputs Z that Z = constant secret X + variable public Y? Let me ask whether SHA1 is designed to be secure for the following case.
You compute each SHA1 of many strings,for example 1 million, where each string is a concatenation of X+Y ,
where X is secret and constant and Y is public and variable.

Thank you in advance.
Answering the question as worded in its body: NO, $\mathrm{SHA1}$ is not designed so that the proposed construction is secure under the stated conditions. The design objective of the $\mathrm{SHA1}$ and $\mathrm{SHA2}$ hashes, as explained by NIST, is that


  it is computationally infeasible to find a message that corresponds to a given message digest, or to find two different messages that produce the same message digest.


and this does not lead to the conclusion that the proposed construction is secure, under any reasonable definition.



In order to answer the (different) question as worded in its title, we need to define "secure". The question is about a construction $F: (X,Y) \to F(X,Y) = \mathrm{SHA1}(X||Y)$. One possible definition of "secure" could be that $Y \to F(X,Y)$ is computationally indistinguishable from a random function for an attacker not knowing $X$ (which implies that $X$ does not leak from examples).

With this wide definition, NO, $\mathrm{SHA1}$ is not secure. It is vulnerable to a length-extension attack, a classical weakness of the Merkle–Damgård construction used in $\mathrm{SHA1}$. An adversary knowing $\mathrm{SHA1}(X||Y)$ can compute $\mathrm{SHA1}(X||Y')$ for some $Y'\ne Y$, on the conditions that $Y'=Y||P||E$; and he knows $E$ and the length of $X||Y$; and $P$ is a specific 65-to-576-bit string depending only on that length. Having many examples of $(Y,\mathrm{SHA1}(X||Y))$ helps the adversary, in that it widens the choices for $Y'$.

To perform this length-extension attack, the adversary can choose $P$ as the 65-to-576-bit string constituting the padding for $X||Y$ when computing $\mathrm{SHA1}(X||Y)$. Then, $\mathrm{SHA1}(X||Y||P||E)$ is computed by starting the normal computation where the block with $E$ starts, with the 160-bit hash state $\mathrm{SHA1}(X||Y)$ rather than the constant specified in $\mathrm{SHA1}$. The attack has very low cost, lower than if the adversary had computed $\mathrm{SHA1}(X||Y')$ directly with knowledge of $X||Y'$.



When the conditions for the length-extension attack are not met (including if the length of $Y$ is fixed, exactly or with a small leeway of +64 bits), or when the objective of the adversary is finding $X$, the construction proposed has considerable cryptanalytic resistance. Up to some limit on the size of $X$, we know no attack much better than brute-forcing $X$. A security argument could be given based on the assumption that the cipher used in the $\mathrm{SHA1}$ round function is secure; but we know that this assumption does not hold at all, if only because of some near-practical attacks on the collision-resistance of $\mathrm{SHA1}$. Therefore I can't quantify the level of resistance better than "AFAIK unbroken for practical purposes", which is not strong enough for a recommendation.



Fortunately, there is a recommendable construct based on $\mathrm{SHA1}$ that is designed to be secure (in the wide definition we considered); this is HMAC with $\mathrm{SHA1}$ as the hash function, $X$ as the key, $Y$ as the message. This construction has a security argument that is not directly harmed by a collision-resistance attack on $\mathrm{SHA1}$. Although the absolute confidence we can have in this argument is the subject of debate and some controversy, it is still very reassuring to me about the security of HMAC, and I am not alone.